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Sequences And Series

Class 11th Mathematics CBSE Solution
Exercise 9.1
  1. an = n (n + 2) Write the first five terms of each of the sequences in whose nth…
  2. a_n = n/n+1 Write the first five terms of each of the sequences in whose nth…
  3. an = 2n Write the first five terms of each of the sequences in whose nth terms…
  4. Write the first five terms of each of the sequences in whose nth terms are a_n =…
  5. a_n = (-1)^n-15^n+1 Write the first five terms of each of the sequences in whose…
  6. a_n = n n^2 + 5/4 Write the first five terms of each of the sequences in whose…
  7. a_n = 4n-3 a_17 , a_24 Find the indicated terms in each of the sequences in…
  8. a_n = n^2/2^n a_7 Find the indicated terms in each of the sequences in whose nth…
  9. an= (-1)n-1 n^3 ; a9 Find the indicated terms in each of the sequences in whose…
  10. a_n = n (n-2)/n+3 a_20 Find the indicated terms in each of the sequences in…
  11. a1 = 3, an = 3an - 1 + 2 for all n 1 Write the first five terms of each of the…
  12. a_1 = - 1 , a_n = a_n-1/n , n geater than or equal to 2 Write the first five…
  13. a1 = a2 = 2, an = an-1 -1, n 2 Write the first five terms of each of the…
  14. The Fibonacci sequence is defined by 1 = a1= a2 and an = an-1 + an - 2, n 2.…
Exercise 9.2
  1. Find the sum of odd integers from 1 to 2001.
  2. Find the sum of all-natural numbers lying between 100 and 1000, which are…
  3. In an A.P., the first term is 2 and the sum of the first five terms is…
  4. How many terms of the A.P. - 6,- 11/2 , - 5, … are needed to give the sum -25?…
  5. In an A.P., if pth term is 1/9 and qth term is 1/p , prove that the sum of first…
  6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find…
  7. Find the sum to n terms of the A.P., whose kth term is 5k + 1.
  8. If the sum of n terms of an A.P. is (pn + qn^2), where p and q are constants,…
  9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n…
  10. If the sum of first p terms of an A.P. is equal to the sum of the first q…
  11. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.…
  12. The ratio of the sums of m and n terms of an A.P. is m^2 : n^2 . Show that the…
  13. If the sum of n terms of an A.P. is 3n^2 + 5n and its mth term is 164, find the…
  14. Insert five numbers between 8 and 26 such that the resulting sequence is an…
  15. If a^n + b^n/a^n-1 + b^n-1 is the A.M. between a and b, then find the value of…
  16. Between 1 and 31, m numbers have been inserted in such a way that the resulting…
  17. A man starts repaying a loan as first instalment of Rs. 100. If he increases…
  18. The difference between any two consecutive interior angles of a polygon is 5°.…
Exercise 9.3
  1. Find the 20th and nth terms of the G.P. 5/2 , 5/4 , 5/8 ,
  2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.…
  3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that…
  4. The 4th term of a G.P. is square of its second term, and the first term is - 3.…
  5. Which term of the following sequences: (A) 2 , 2 root 2 , 4 , l 128? (B) root 3…
  6. For what values of x, the numbers - 2/7 , x , - 7/2 are in G.P.?
  7. 0.15, 0.015, 0.0015,... 20 terms. Find the sum to indicated number of terms in…
  8. root 7 root 21 , 3 root 7 , n terms. Find the sum to indicated number of terms…
  9. 1, - a, a^2 , - a, ... n terms (if a plus or minus - 1). Find the sum to…
  10. x^3 , x5, x^7 , ... n terms (if x plus or minus i 1). Find the sum to indicated…
  11. Evaluate sum _ k = 1^11 (2+3^k)
  12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find…
  13. How many terms of G.P. 3, 3^2 , 3^3 , … are needed to give the sum 120?…
  14. The sum of first three terms of a G.P. is 16 and the sum of the next three…
  15. Given a G.P. with a = 729 and 7th term 64, determine S7.
  16. Find a G.P. for which sum of the first two terms is - 4 and the fifth term is 4…
  17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove…
  18. Find the sum to n terms of the sequence 8, 88, 888, 8888… .
  19. Find the sum of the products of the corresponding terms of the sequences 2, 4,…
  20. Show that the products of the corresponding terms of the sequences a, ar, ar^2…
  21. Find four numbers forming a geometric progression in which the third term is…
  22. If the pth, qth and rthterms of a G.P are a, b and c, respectively. Prove that…
  23. If the first and the nth term of a G.P. are a and b, respectively, and if P is…
  24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms…
  25. If a, b, c and d are in G.P. show that (a^2 + b^2 + c^2) (b^2 + c^2 + d^2) =…
  26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.…
  27. Find the value of n so that a^n+1 + b^n+1/a^n + b^n may be the geometric mean…
  28. The sum of two numbers is 6 times their geometric mean, show that numbers are…
  29. If A and G be A.M. and G.M., respectively between two positive numbers, prove…
  30. The number of bacteria in a certain culture doubles every hour. If there were…
  31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays…
  32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively,…
Miscellaneous Exercise
  1. Show that the sum of (m + n)th and (m - n)th terms of an A.P. is equal to twice…
  2. If the sum of three numbers in A.P., is 24 and their product is 440, find the…
  3. Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show…
  4. Find the sum of all numbers between 200 and 400 which are divisible by 7.…
  5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.…
  6. Find the sum of all two-digit numbers which when divided by 4, yields 1 as…
  7. If f is a function satisfying f (x + y) = f(x) f(y) for all v ucnt such that…
  8. The sum of some terms of G.P. is 316 whose first term and the common ratio are 6…
  9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90.…
  10. The sum of three numbers in G.P. is 66. If we subtract 1, 7, 21 from these…
  11. A G.P. consists of an even number of terms. If the sum of all the terms is 5…
  12. The sum of the first four terms of an A.P. is 66. The sum of the last four…
  13. If a+bx/a-bx = b+cx/b-cx = c+dx/c-dx (x not equal 0) , then show that a, b, c…
  14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a…
  15. The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that (q -…
  16. If a (1/h + 1/r) , b (1/r + 1/a) , c (1/a + 1/h) are in A.P., prove that a, b,…
  17. If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in…
  18. If a and b are the roots of x^2 - 3x + p = 0 and c, d are roots of x^2 - 12x +…
  19. The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show…
  20. If a, b, c are in A.P.; b, c, d are in G.P. and 1/n , 1/d , 1/a are in A.P.…
  21. 6 + 66 + 666 + … Find the sum of the following series up to n terms:…
  22. 6 + . 66 + . 666 + … Find the sum of the following series up to n terms:…
  23. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.…
  24. Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + ……
  25. If S1, S2, S3 are the sum of first n natural numbers, their squares and their…
  26. Find the sum of the following series up to n terms:
  27. Show that 1 x 2^2 + 2 x 3^2 + l n x (n+1)^2/1^2 x 2+2^2 x 3 + l n^2 x (n+1) =…
  28. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to…
  29. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to…
  30. A person writes a letter to four of his friends. He asks each one of them to…
  31. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.…
  32. A manufacturer reckons that the value of a machine, which costs him Rs. 15625,…
  33. 150 workers were engaged to finish a job in a certain number of days. 4 workers…

Exercise 9.1
Question 1.

Write the first five terms of each of the sequences in whose nth terms are:

an = n (n + 2)


Answer:

The nth term of the sequence is given as an = n (n + 2)


The first five terms of the sequence will be given by substituting the value by n as 1, 2, 3, 4 and 5


The sequence so obtained now is


a1= 1(1 + 2) = 3


a2 = 2(2 + 2) = 8


a3 = 3(3 + 2) = 15


a4= 4(4 + 2) = 24


a5= 5(5 + 2) = 35


The required first five numbers of the sequence are 3, 8, 15, 24, 35



Question 2.

Write the first five terms of each of the sequences in whose nth terms are:



Answer:

The given nth term of the sequence is


The first five terms of the sequence would be given by substituting the value if n as 1, 2, 3, 4, 5


The required terms would be







The first five numbers of sequence so obtained would be



Question 3.

Write the first five terms of each of the sequences in whose nth terms are:

an = 2n


Answer:

The nth term of the sequence is an = 2n


The first five terms of the sequence would be obtained by putting value of n as 1, 2, 3, 4, 5


a1= 21 = 2


a2 = 22= 4


a3= 23 = 8


a4= 24=16


a5= 25 = 32


The required first five numbers of the sequence are 2, 4, 8, 16 and 32



Question 4.

Write the first five terms of each of the sequences in whose nth terms are


Answer:

The nth term of the sequence is given as


The first five terms of the sequence would be obtained by putting the value of n as 1, 2, 3, 4 and 5







The first five terms of the sequence are


Question 5.

Write the first five terms of each of the sequences in whose nth terms are:



Answer:

The nth term of the sequence is given as an= (-1)n-1 5n + 1


The first five terms of the sequence would be given by putting values of n as 1, 2, 3, 4 and 5


a1= (-1)1-1 51 + 1 = (-1)0 52 = 25 (∵ a0 =1)


a2 = (-1)2-1 52 + 1 = (-1)1 53 = -125


a3 = (-1) 3-1 53 + 1 = (-1)2 54 = 625


a4 = (-1)4-1 54 + 1 = (-1)3 55 = -3125


a5= (-1)5-1 55 + 1 = (-1)4 56 = 15625


∴ the first five terms of the sequence are 25, -125, 625, -3125 and 15625



Question 6.

Write the first five terms of each of the sequences in whose nth terms are:



Answer:

The nth term of the given sequence is



The first five terms of the sequence would be given by putting the value of n as 1, 2, 3, 4 and 5








The first five terms of the sequence are



Question 7.

Find the indicated terms in each of the sequences in whose nth terms are:



Answer:

The nth term of the sequence is

an = 4n – 3…….1


We have to find out a17 and a24


Putting value of n = 17 and n = 24 in the given expression 1


a17 = 4(17) – 3 = 65


And a24 = 4(24) – 3 = 93



Question 8.

Find the indicated terms in each of the sequences in whose nth terms are:



Answer:

The given nth term of the sequence is


………….1


We have to find out a7, so putting value of n = 7 in given expression 1




Question 9.

Find the indicated terms in each of the sequences in whose nth terms are:

an= (-1)n-1 n3 ; a9


Answer:

The given nth term of the sequence is an= (-1)n-1 n3


We have to find out value of a9, putting value of n = 9 in given expression


a9 = (-1)8 93 = 729



Question 10.

Find the indicated terms in each of the sequences in whose nth terms are:



Answer:

In the given expression the value of nth term is


We have to find out value of a20 , putting the value of n = 20 in given expression




Question 11.

Write the first five terms of each of the sequences in and obtain the corresponding series:

a1 = 3, an = 3an – 1 + 2 for all n > 1


Answer:

Given here are a1 = 3,

an = 3a(n – 1) + 2…………1


To find out first five terms of the sequence we will put value of n = 2, 3, 4, 5 in 1


a2= 3(3)2-1 + 2 = 3 × 3 + 2 = 11


a3 = 3a2 + 2 = 3 × 11 + 2 = 35 (∵ an-1 = a3-1 = a2 )


a4 = 3a3 + 2 = 105 + 2 = 107


a5 = 3a4 + 2 = 3 × 107 + 2 = 323


Hence the first five terms of the sequence are 3, 11, 35, 107, 323


And the corresponding sequence is 3 + 11 + 35 + 107 + 323 + …….



Question 12.

Write the first five terms of each of the sequences in and obtain the corresponding series:



Answer:

Here given a1= -1 and


………………1


To find out the first five terms of the sequence we will put


n = 2, 3, 4, 5 in the expression 1






Hence the first five terms of the sequence are


The corresponding series is




Question 13.

Write the first five terms of each of the sequences in and obtain the corresponding series:

a1 = a2 = 2, an = an-1 -1, n > 2


Answer:

Here given, a1 = a2 = 2, an = an-1 -1, n > 2


a3 = a2 -1 = 2-1 = 1


a4 = a3 -1 = 1-1 = 0


a5 = a4 -1 = 0 -1 = -1


The first five numbers of the sequence are 2, 2, 1, 0 ,-1


The corresponding series is 2 + 2 + 1 + 0 + (-1) + ………



Question 14.

The Fibonacci sequence is defined by

1 = a1= a2 and an = an-1 + an – 2, n > 2.

Find, for n = 1, 2, 3, 4, 5


Answer:

Here given a1 = a2 =1

And an = an-1 + an-2 , n > 2.


a3 = a2 + a1 = 1 + 1 = 2


a4 = a3 + a2 = 2 + 1 = 3


a5 = a4 + a3 = 3 + 2 = 5


a6 = a5 + a4 = 5 + 3 = 8


Now putting the value of n = 1, 2, 3, 4, 5 in


For








Exercise 9.2
Question 1.

Find the sum of odd integers from 1 to 2001.


Answer:

The odd integers from 1 to 2001 are 1, 3, 5 …1999, 2001.


This sequence forms a Arithmetic Progression


Let the first term be ‘a’ and common difference ‘d’.


Let n be the total number of terms in the series.


Here, first term, a = 1


Common difference, d = 2


If l denotes the last term of the series


Then, l = a + (n – 1) × d


Here, l = 2001


⇒ 2001 = 1 + (n – 1) × 2


⇒ 2001 – 1 = (n – 1) × 2


⇒ 2000/2 = n – 1


⇒ 1000 + 1 = n


∴ n = 1001


Sum of A.P. =




⇒ Sn = 1002001


∴ The sum of odd numbers from 1 to 2001 is 1002001.



Question 2.

Find the sum of all-natural numbers lying between 100 and 1000, which are multiples of 5.


Answer:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.


Let the first term be ‘a’ and common difference ‘d’.


Let n be the total number of terms in the series.


Here, first term, a = 105


Common difference, d = 5


If l denotes the last term of the series


Then, l = a + (n – 1) × d


Here, l = 995


⇒ 995 = 105 + (n – 1) × 5


⇒ 995 – 105 = (n – 1) × 5


⇒ 890/5 = n – 1


⇒ 178 + 1 = n


∴ n = 179


Sum of A.P. =




⇒ Sn = 179 × 550 = 98,450.



Question 3.

In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.


Answer:

Given, first term a = 2


Let d be the common difference


Let Sn denote sum of n terms,



tn denote nth term.


tn = a + (n – 1)d


S1 = Sum of first five terms.


S2 = Sum of next five terms.



S1 = 10 × (1 + d)


Sum of next 10 terms = Sum of first 10 terms - Sum of first 5 terms


S2 = 20 + 45d – 10 – 10d = 10 + 35d


Also, S1 = 1/4 S2


⇒ 10 + 10d = 1/4 (10 + 35d)


⇒ 40 + 40d = 10 + 35d


⇒ 30 = -5d


⇒ d = -6


t20 = 2 + (20 – 1) × (-6)


t20 = 2 – 19 × 6 = 2 – 114


t20 = -112


Question 4.

How many terms of the A.P. – 6,-, – 5, … are needed to give the sum –25?


Answer:

First term a = -6


Second term = -11/2


Let d be the common difference.


d = Second term – First term


d = -11/2 – (-6) = 6 – 11/2 = 1/2


Sn = Sum of n terms of AP = -25









⇒ n2 – 25n + 100 = 0


⇒ n2 – 20n – 5n + 100 = 0


⇒ n × (n – 20) – 5 × (n – 20) = 0


⇒ (n – 20) × (n – 5) = 0


⇒ n = 20 or 5


Question 5.

In an A.P., if pth term is and qth term is, prove that the sum of first pq terms is 1/2 (pq +1), where p q.


Answer:

Let tn = nth tern of AP


tn = a + (n – 1)d


Sn = sum of n terms of AP



Let a be the first term and d common difference.


Given, in an AP


pth term = 1/q


………………………………..(I)


qth term = 1/p


………………………………(II)


(I) – (II)





Putting value of d in (I)








Thus, the sum of the first pq terms of the A.P. is .


Question 6.

If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.


Answer:

Given terms of A.P. 25, 22, 19, …………..


Let the sum of n terms of the given A.P. be 116.


Sn = 116


First Term = a = 25, Common Difference = d = 22 – 19 = -3



Putting the value of a, d and Sn



⇒ 116 × 2 = n [50 – 3n + 3]


⇒ 232 = 53n – 3n2⇒ 3n2 – 53n + 232 = 0


⇒ 3n2 – 24n – 29n + 232 = 0


⇒ 3n (n – 8) – 29 (n – 8) = 0


⇒ (3n – 29) (n – 8) = 0


⇒ n = 8 or 29/3


n cannot be equal to 29/3. Therefore, n = 8


Last term = a8 = a + (n – 1)d


a8 = 25 + (8 – 1) × (-3) = 25 – 21


∴ a8 = 4



Question 7.

Find the sum to n terms of the A.P., whose kth term is 5k + 1.


Answer:

Let the first term be a and common difference be d.


Given that kth term of the A.P. is 5k + 1.


kth term = ak = a + (k – 1)


∴ a + (k – 1)d = 5k + 1 ⇒ d k + (a - d) = 5k + 1


Comparing the coefficient of k, we obtain d = 5 and a – d = 1


⇒ a – 5 = 1


⇒ a = 6



Putting the value of a and d





Question 8.

If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.


Answer:

Given: Sum of n terms of A.P. = pn + qn2

We know,

where, n = nth term
a = first term and
d = common difference
Now equating the given Sum with the formula we get,



Comparing the coefficients of n2 on both sides, we obtain



⇒ d = 2q


∴ the common difference of the A.P. is 2q.


Question 9.

The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.


Answer:

Let a1, a2, and d1, d2 be the first terms and the common difference of the first and second arithmetic progression respectively.


Given:
We know that Sum of n terms of an A.P is given by,

where, n = number of terms, a = first term and d = common difference of A.P.

So now we have,



Now as we need to find the ratio of first 17 terms, we want the numerator and denominator of the form (a + 17 d)
2 a1 + (n - 1) d1 = a + 17 d
So we need to multiply R.H.S by 2 and we get,
2 a1 + (n - 1) d = 2 a + 34 d
Equating the terms we get,
n - 1 = 34, n = 35.


Putting n = 35 in the above equation






Thus, the ratio of 18th term of both the A.P.s is 179: 321


Question 10.

If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.


Answer:

Let a and d be the first term and the common difference of the A.P. respectively.


Given,


Sum of first p terms =


Sum of first q terms =


Sp = Sq



⇒ p [2a + pd – d] = q [2a + qd – d]


⇒ 2ap + p (p – 1)d = 2aq + q (q – 1)d


⇒ 2a (p – q) + d [p(p – 1) – q(q – 1)] = 0


⇒ 2a (p – q) + d [p2 – q2 – (p – q)] = 0


⇒ 2a (p – q) + d [(p + q)(p – q) – (p – q)] = 0


⇒ (p – q) [2a + d (p + q – 1)] = 0


⇒ [ 2a + d (p + q – 1)] = 0





⇒ Sp+q = 0



Question 11.

Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that:


Answer:

Given:Sp = a, Sq = b, and Sr = c

To Prove:

Proof:

Let a1 be the first terms and d be the common difference of the A.P.

We know that sum of n terms of an A.P is given by:


where, a = first term
n = nth terms
d = common difference

Therefore, we have,

Sum of first p terms =


......(1)


Sum of first q terms =


......(2)


Sum of first p terms =


.....(3)


Subtracting (2) from (1)





Subtracting (3) from (2)





From (IV) and (V),



⇒ pq (p – q) (2br – 2cq) = qr (q – r) (2aq – 2bp)


⇒ p (p – q) (2br – 2cq) = r (q – r) (2aq – 2bp)


⇒ (aqr – bpr) (q – r) = (bpr – cpq) (p – q)


Dividing both sides by pqr





Hence, proved.


Question 12.

The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).


Answer:

Let a and d be the first term and common difference of the A.P.


Given,


Sum of m terms of A.P. = Sm



Sum of n terms of A.P. = Sn






Putting m = 2m – 1 and n = 2n – 1 in the above equation






∴ Ratio of mth and nth term is (2m – 1) : (2n – 1).



Question 13.

If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.


Answer:

Let a and d be the first term and common difference of A.P.


Given, Sum of n terms of A.P. = Sn = 3n2 + 5n ………….(I)



mth term of A.P. = tm = 164 ……………………….(III)


⇒ tm = a + (m – 1)d …………………..(IV)


Equating (I) and (II)



⇒ 2an + n2d – dn = 6n2 + 5n


⇒ (2a – d)n + n2d = 6n2 + 10n


Comparing the coefficients of n2, we get


d = 6


Comparing the coefficients of n, we get


2a – d = 10


Putting the value d = 6


⇒ 2a – 6 = 10 ⇒ 2a = 10 + 6


⇒ 2a = 16


⇒ a = 8


mth term = tm = 164 = 8 + (m – 1)6


⇒ 164 – 8 = (m – 1)6


⇒ m – 1 = 156/6


⇒ m – 1 = 26


⇒ m = 26 + 1


∴ m = 27



Question 14.

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.


Answer:

Let A1, A2, A3, A4, and A5 be five numbers between 8 and 26


such that 8, A1, A2, A3, A4, A5, 26 is an A.P.


Here, First term = a = 8,


Last Term = b = 26,


Total no. of terms = n = 7


Therefore, 26 = 8 + (7 – 1) d


⇒ 6d = 26 – 8 = 18


⇒ d = 3


A1 = a + d = 8 + 3 = 11


A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14


A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17


A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20


A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23


Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.



Question 15.

If is the A.M. between a and b, then find the value of n.


Answer:

A.M. of a and b =


Given, is the A.M. between a and b.



⇒ (a + b) (an-1+ bn-1) = 2an + 2bn


⇒ an + abn-1 + ban-1 + bn = 2an + 2bn


⇒ abn-1 + ban-1 = an + bn


⇒ abn-1 – bn = an – ban-1


⇒ bn-1 (a – b) = an-1 (a – b)




⇒ n – 1 = 0


∴ n = 1



Question 16.

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.


Answer:

Let A1, A2, … Am be m numbers such that 1, A1, A2, … Am, 31 is an A.P.


Here, First term = a = 1,


Last term = b = 31,


Total no. of terms = n = m + 2


∴ 31 = 1 + (m + 2 – 1) d


⇒ 30 = (m + 1)d



A1 = a + d


A2 = a + 2d


A3 = a + 3d …


∴ A7 = a + 7d

It is because here the 7th term is between 1 and 31. So counting from 1 the 7th term will be the 8th term.


Am-1 = a + (m – 1)d


Also given,







⇒ 9 (m + 211) = 5 (31m – 29)


⇒ 9m + 1899 = 155m - 145


⇒ 155m – 9m = 1899 + 145


⇒ 146m = 2044


∴ m = 14


Question 17.

A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment?


Answer:

First instalment = Rs 100


Instalment increases by Rs 5


Second Instalment = Rs 105


So, every month instalment Increases by 5


Man repays 100, 105, 110, ………… every month


This forms an A.P.


A.P. is 100, 105, 110, 105, …….


First term of A.P. = 100


Common difference = 5


nth term of A.P. = tn = a + (n – 1)d


30th term = t30 = 100 + (30 – 1) × 5


= 100 + 29 × 5 = 100 + 145


t30 = 245


Man will pay Rs 245 in 30th instalment.



Question 18.

The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.


Answer:

Smallest Angle = 120°


Difference between any two consecutive interior angles of a polygon = 5°


The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°.


We know, sum of all angles of a polygon with n sides = 180° (n – 2)


Sn = 180° (n – 2)



Equating both we get




⇒ n (240 + 5n – 5) = 360n – 720


⇒ 5n2 + 240n – 5n – 360n + 720 = 0


⇒ 5n2 - 125n + 720 = 0


⇒ n2 – 25n + 144 = 0


⇒ n2 – 16n – 9n + 144 = 0


⇒ n (n – 16) – 9 (n – 16) = 0


⇒ (n – 9) (n – 16) = 0


∴ n = 9 or 16



Exercise 9.3
Question 1.

Find the 20th and nth terms of the G.P.


Answer:

Given: G.P.


We know that in G.P an = arn-1


Here, n: number of terms


a: First term = 5/2


r: common ratio


Common ratio of the given G.P is: (∴ r = 1/2)


Here, nth team of the G.P :



∴ nth term of the given G.P. is


20th term of the given G.P. is


∴ nth and 20th term of the given G.P. are: respectively.



Question 2.

Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.


Answer:

Given: 8th of the G.P. is 192 and common ratio is 2.


That is,


a8 = 192 and r = 2.


We know that in G.P an = arn-1


Here, n: number of terms


a: First term = 5/2


r: common ratio


∴ a8 = a(2)8-1 (∵ r = 2 and n = 8)


⇒ 192 = a(2)7


⇒ a × 128 = 192



Now,



∴ a12 = 3072



Question 3.

The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.


Answer:

Given: 5th, 8th and 11th terms of a G.P. are p, q and s, respectively


We know that in G.P an = arn-1


Here, n: number of terms


a: First term


r: common ratio


Here,


a5 = ar5-1 = ar4


⇒ p = ar4 (∵ 5th term of G.P. is given p) –1


Similarly,


a8 = ar8-1 = ar7


⇒ q = ar7 (∵ 7th term of G.P. is given q) –2


a11 = ar11-1 = ar10


⇒ s = ar10 (∵ 11th term of G.P. is given s) –3


We can observe that:


q × q = p × s (that is, ar7 × ar7 = ar4 × ar10)


∴ q2 = ps


Hence proved



Question 4.

The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.


Answer:

Given: a4 = (a2)2 and a = –3


We know that in G.P an = arn-1


∴ a2 = (–3) × r2-1 = –3r(∵ a = –3)


Similarly,


a4 = –3r3


∴ –3r3 = (-3r)2 (∵ a4 = (a2)2 )


∴ –3r3 = 9r2


⇒ r = –3


∴ r = –3


Now,


a7 = ar7-1


⇒ a7 = (–3)( –3)6 (∵ a = –3 and r = –3)


⇒ a7 = ( –3)7 = – 2187.


∴ a7 = – 2187



Question 5.

Which term of the following sequences:

(A)

(B)

(c)


Answer:

A)


Given: 2, 2√2, 4….. and an = 128


Here, in the above G.P.


a = 2


Common ratio(r) =


We know that in G.P an = arn-1


∴ an = 2 × (√2)n-1


⇒ 128 = 2 × (√2)n-1


⇒ (√2)n-1 = 128/2 = 64


= 64 (∵ √x = x1/2)


Apply ln on both sides


We get






⇒ n – 1 = 12


⇒ n = 13


∴ a13 = 128


B)


Given: √3, 3, 3√3 ….. and an = 729


Here, in the above G.P.


a = √3


Common ratio(r)


We know that in G.P an = arn-1


∴ an = √3 × (√3)n-1


⇒ 729 = (√2)n


⇒ (√2)n = 729


= 729 (∵ √x = )


Apply ln on both sides


We get






⇒ n = 12


∴ a12 = 729


C)


Given: and an =


Here, in the above G.P.


a =


Common ratio(r) = =


We know that in G.P an = arn-1


∴ an =





Apply ln on both sides


We get





⇒ n = 9


∴ a9 =



Question 6.

For what values of x, the numbers are in G.P.?


Answer:

Given:


For the given Sequence to be in G.P. Common ratio between two adjacent numbers in the sequence should be equal.


That is: Common ratio between = Common ratio between




⇒ (1)2 = x2


⇒ x =


⇒ x =


∴ For the given sequence to be in G.P. x should be .



Question 7.

Find the sum to indicated number of terms in each of the geometric progressions in

0.15, 0.015, 0.0015,... 20 terms.


Answer:

Given: 0.15, 0.015, 0.0015,... 20 terms.


Sum of n terms of a G.P. is given by: (a: First term of G.P, r: common difference of G.P, n: Number of terms of the G.P)


First term of the Given G.P (a) = 0.15


Common difference of the given G.P(r) = = = 0.1


Number of terms(n): 5


Let the sum of 20 terms be s






∴ The sum of 20 terms of the given sequence is: [1-(0.1)20]



Question 8.

Find the sum to indicated number of terms in each of the geometric progressions in

terms.


Answer:

Given: √7, √21, 3√7 ,... n terms.

Sum of n terms of a G.P. is given by: (a: First term of G.P, r: common difference of G.P, n: Number of terms of the G.P)

First term of the Given G.P (a) = √7

Common difference of the given G.P(r) = = √3

Number of terms(n): n

Let the sum of n terms be s

Now, as r > 1,



Question 9.

Find the sum to indicated number of terms in each of the geometric progressions in

1, – a, a2, – a, ... n terms (if a – 1).


Answer:

Given: 1, – a, a2, – a, ... n terms


Sum of n terms of a G.P. is given by: (a: First term of G.P, r: common difference of G.P, n: Number of terms of the G.P)


First term of the Given G.P (a) = 1


Common difference of the given G.P(r) = = -a


Number of terms(n): n


Let the sum of n terms be s


∴ s =


⇒ s =


⇒ s = [


∴ The sum of n terms of the given sequence is: [1-(-a)n]



Question 10.

Find the sum to indicated number of terms in each of the geometric progressions in

x3, x5, x7, ... n terms (if x 1).


Answer:

Given: x3, x5, x7, ... n terms


Sum of n terms of a G.P. is given by: (a: First term of G.P, r: common difference of G.P, n: Number of terms of the G.P)


First term of the Given G.P (a) = x3


Common difference of the given G.P(r) = = x2


Number of terms(n): n


Let the sum of n terms be s




∴ The sum of n terms of the given sequence is:



Question 11.

Evaluate


Answer:

Given:



Here, we can see that is in G.P.


Where a = 3, r = 3


We know that Sum of terms in G.P. is given by : (here ‘n’ is the number of terms)







Question 12.

The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.


Answer:

Given: Sum of first three terms of a G.P. is and their product is 1

Let , a , ar be the three terms in G.P.

Given: The product of three terms = 1


⇒ a3 = 1

a = 1

Also,


10r2 - 29r + 10 = 0

10r2 - 25r - 4r + 10 = 0

5r(2r - 5) - 2(2r - 5) = 0



Question 13.

How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?


Answer:

Given: G.P. 3, 32, 33, … and Sum = 120


Here a = 3 and r = = 3


The Sum of terms in G.P. is given by:


= 120


= 120


= 120(2)


= 240


= = 80


⇒ 3n = 80 +1 = 81


⇒ 3n = 81


Applying ln on both sides, we get





⇒ n = 4



Question 14.

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.


Answer:

Given: The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.


Let a, ar, ar2, ar3, ar4, ar5 be the terms of the G.P


Here a + ar +ar2 = 16 (∵ given that sum of first 3 terms is 16) -------–1


Also, ar3, ar4, ar5 = 128 (∵ given that sum of next 3 terms is 128)--------------–2


Divide eq –2 and eq –1


We get,




⇒ r3 = 8


⇒ r = ∛8


⇒ r = 2


Here a + ar + ar2 = 16


⇒ a × (1 + r + r2) = 16


⇒ a × (1 + 2 + (2)2) = 16


⇒ a × (1 + 2 + 4) = 16


⇒ a × (7) = 16


⇒ a =


The Sum of terms in G.P. is given by:


=


∴ First term of the G.P is , common ratio of the G.P is 2 and Sum of n terms of the G.P is



Question 15.

Given a G.P. with a = 729 and 7th term 64, determine S7.


Answer:

Given: a = 729 and a7 = 64.


nth term of a G.P is given by arn-1


∴ a7 = ar7-1


⇒ 64 = 729 × (r)6


⇒ r6 =


⇒ r6 =


∴ r =


The Sum of terms in G.P. is given by: (∵ r < 1)



⇒ 37 – 27 = 2187 – 128 = 2059


∴ S7 = 2059



Question 16.

Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.


Answer:

Given: S2 = —4 and a5 = 4 × a3


The Sum of terms in G.P. is given by:


∴ S2 = = —4


= —4


⇒ a(1+r) = —4 –1


Here,


a5 = 4 × a3


⇒ ar4 = 4 × ar2 (∵ nth term of the G.P is arn-1)


⇒ r2 = 4


⇒ r = 2


∴ r = +2 or –2


Case 1: r = +2


From eq –1


a(1+r) = —4


⇒ a(1+2 ) = —4


⇒ a =


∴ G.P :


Case 2: r = –2


From eq –1


a(1+r) = —4


⇒ a(1+(–2) ) = —4


⇒ a = = 4


∴ G.P : 4 , –8, 16, –32,…….


∴ The possible G.P ‘s are or 4 , –8, 16, –32,…….



Question 17.

If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.


Answer:

Given: 4th, 10th and 16th terms of a G.P. are x, y and z, respectively


Let a be the first term and r be the common ratio of the G.P.


Here,


We know that nth term of the G.P is given by arn-1



a4 = ar3 = x —1


a10 = ar9 = y —2


a16 = ar15 = z —3


Dividing eq-(2) by eq-(1), we obtain




Dividing eq(3) by eq-(2), we obtain




That is


Thus, x, y, z are in G. P



Question 18.

Find the sum to n terms of the sequence 8, 88, 888, 8888… .


Answer:

Given: sequence: 8, 88, 888, 8888… .


Sn = 8 + 88 + 888 + 8888 + …….


⇒ Sn = 8[1 + 11+ 111 + 1111 +…….]


⇒ Sn = × [9 + 99+ 999 + 9999 +…….]


⇒ Sn = × [(10 – 1) + (102 – 1)+ (103 – 1) + (104 – 1) +…….]


⇒ Sn = × [10 + 102 + 103 + 104 +…….] – × [1 + 1+ 1 +1 +…….]


⇒ Sn = × [10 + 102 + 103 + 104 +…….] – × [1 + 1+ 1 +1 +…….]


⇒Sn = (∵ sum of n terms of G.P is , here a = 10 and r = 10)


⇒Sn =


⇒Sn =


⇒Sn =




Question 19.

Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,16, 32 and 128, 32, 8, 2,


Answer:

Given: the sequences are 2, 4, 8,16, 32 and 128, 32, 8, 2,


The Required answer is: 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 ×


⇒ 64 × [4 + 2 + 1 + + ]


Here the terms are in G.P


Where a = 4, r = and n = 5


The Sum of terms in G.P. is given by:


∴ S = 64 ×


⇒ S = 64 ×


⇒ S = 64 ×


⇒ S = 2 ×


⇒ S = = 496


∴ The required answer is 496.



Question 20.

Show that the products of the corresponding terms of the sequences a, ar, ar2,…arn – 1 and A, AR, AR2, … ARn – 1 form a G.P, and find the common ratio.


Answer:

Given: The sequences a, ar, ar2,…arn – 1 and A, AR, AR2, … ARn – 1


The products of the corresponding terms of the G.P is


a × A, ar × AR, ar2 × AR2, ……….., arn × ARn


Here,


= rR


= rR


∴ The product of the corresponding terms of the given sequence forms a G.P and the common ratio is rR



Question 21.

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.


Answer:

Given: a3 = a1 + 9 and a2 = a4 + 18


Let a1 = a, a2 = ar, a3 = ar2, a4 = ar3


Here,


a3 = a1 + 9


⇒ ar2 = a + 9


⇒ ar2 – a = 9


⇒ a(r2 – 1) = 9 – 1


and,


a2 = a4 + 18


⇒ ar= ar3 + 18


⇒ ar – ar3 = 18


⇒ ar(1 – r2) = 18 – 2


Divide eq –2 by eq –1





⇒ r = – 2


Substitute r in eq—1


a × ((-2)2 – 1) = 9


a × (4 – 1) = 9


a × (3) = 9


a = = 3


∴ G.P is : 3 , 3(– 2), 3(– 2)2, 3(– 2)3


⇒ 3, —6 , 12 , —24


∴ G.P is 3, —6, 12, —24


Question 22.

If the pth, qth and rthterms of a G.P are a, b and c, respectively. Prove that

aq – r × br – p × cP – q = 1.


Answer:

Given pth, qth and rthterms of a G.P are a, b and c, respectively


Here


ap = a = arp-1


aq = b = arq-1


ar = c = arr-1


Now,


aq – r × br – p × cP – q = (arp - 1)q - r × (arq - 1)r - p × (arr-1)p – q


⇒ aq – r × br – p × cP – q = (a(q - r) × r(p – 1)(q-r)) × (a(r - p) × r(q – 1)(r - p)) × (a(p - q) × r(r – 1)(p - q))


⇒ aq – r × br – p × cP – q = (a(q – r) × r(pq – q - pr +r)) × (a(r - p) × r(qr – r - pq + p )) × (a(p - q) × r(pr – p – qr + q))


⇒ aq – r × br – p × cP – q = (a(q – r + r – p + p - q) × r(pq – q – pr +r + qr - r –pq + p +pr – p –qr + q))


⇒ aq – r × br – p × cP – q = (a0 × r0) = 1


∴ aq – r × br – p × cP – q = 1


Hence proved.



Question 23.

If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.


Answer:

Given: Let the first and the nth term of a G.P. be a and b, respectively, and P be the product of n terms.


Here,


a1 = a = a


an = b = arn-1 —1


Here,


P = Product of n terms


⇒ P = (a) × (ar) × (ar2) × ….. × (arn-1)


⇒ P = (a × a × …a) × (r × r2 × …rn-1)


⇒ P = an × r 1 + 2 +…(n–1) —2


Here,


1, 2, …(n – 1) is an A.P.


The sum of n terms of an A.P is given by : (here n: no of terms, a:first term, d:common difference)


∴ 1+2+3+….+(n-1) =


∴ 1+2+3+….+(n-1) =


∴ P = an × r 1 + 2 +…+(n–1)


⇒ P = an ×


⇒ P2 =


⇒ P2 =


⇒ P2 =


⇒ P2 =


⇒ P2 = from eq –1


∴ P2 = (ab)n



Question 24.

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from

(n + 1)th to (2n)th term is


Answer:

Let a be the first term and r be the common ratio of the G.P.


Sum of n terms =


Since there are n terms from (n +1)th to (2n)th term,


Sum of terms from(n + 1)th to (2n)th term =


Here,


an+1 = arn+1–1 = arn


∴ Required ratio =


∴ The ratio of the sum of first n terms of a G.P. to the sum of terms from


(n +1)th to (2n)th term is = .



Question 25.

If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2.


Answer:

Given: a, b, c and d are in G.P.


Here,


a, b, c, d are in G.P.


bc = ad — (1)


b2 = ac —(2)


c2 = bd –(3)


we have to be prove that,


(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2


From R.H.S.


(ab + bc + cd)2


= (ab + ad + cd)2 (From eq–1)


= (ab + d (a + c))2


= a2b2 + 2abd (a + c) + d2 (a + c)2


= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)


= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 (from eq– 1 and eq—2)


= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2


= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2


From eq – 2 and eq – 3


= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)


= (a2 + b2 + c2) (b2 + c2 + d2)


= L.H.S.


∴ L.H.S. = R.H.S.


(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2


Question 26.

Insert two numbers between 3 and 81 so that the resulting sequence is G.P.


Answer:

Let a1 and a2 be two numbers between 3 and 81 such that the series, 3, a1, a2, 81, forms a G.P.


Let a0 be the first term and r be the common ratio of the G.P.


∴81 = ar3


⇒81 = (3)r3


r3 = 27


r = 3


a1 = a0r = (3) (3) = 9


a2 = a0 r2 = (3) (3)2 = 27


∴ The required two numbers are 9 and 27.



Question 27.

Find the value of n so that may be the geometric mean between a and b.


Answer:

G.M of two numbers in G.P. is given by √ab


= √ab


By squaring on both sides we get,


= ab









⇒ 2n+1 = 0


⇒ 2n = –1


⇒ n =


∴ Value of n for to be G.M is



Question 28.

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .


Answer:

Let the two numbers be a and b.


G.M. = √ab


According to the given condition,


a + b = 6√ab


squaring on both sides we get,


⇒ (a + b)2 = 36(ab) —1


Here,


(a—b)2 = (a + b)2 – 4ab = 36ab – 4ab = 32ab


⇒ a—b = √32√ab = 4√2√ab —2


Adding eq(1) and eq(2), we get


2a = (6 + 4√2)√ab


⇒ a = (3 + 2√2)√ab


Substituting the value of a in eq(1), we obtain


b = 6√ab – (3 + 2√2)√ab


⇒ b = (3 — 2√2)√ab


Now,



∴ The required ratio is



Question 29.

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .


Answer:

Given: A and G are A.M. and G.M. between two positive numbers.


Let the two numbers be a and b.


∴ AM = A = —1


GM = G = √ab —2


From eq-1 and eq-2, we get


a + b = 2A —3


ab = G2 —4


Substituting the value of a and b from eq-3 and eq-4 in


(ab)2 = (a + b)2 – 4ab, we get


(ab)2 = 4A2 – 4G2 = 4 (A2G2)


(ab)2 = 4 (A + G) (AG)


(a – b) = 2 —5


From eq-3 and eq-5, we get


2a = 2A + 2


⇒ a = A+2


Substituting the value of a in eq-3, we get


b = 2A – A - = A –


Thus, the two numbers are .



Question 30.

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?


Answer:

Given: there were 30 bacteria present originally and culture doubles every hour.


∴ a0 = 30


∴ After an hour culture is a0 × 2 = 30 × 2 = 60


Since the count doubles every hour it forms a G.P. with r = 2


Let the G.P be a0,a1r,a2r2........


⇒ The culture at the end of 2nd hour = ar2 = 30 × 22 = 120


⇒ The culture at the end of 4th hour = ar4 = 30 × 24 = 480


⇒ The culture at the end of nth hour = arn = 30 × 2n


∴ Culture at end pf 2nd , 4th , nth hours are 120, 480, 30 × 2n respectively.



Question 31.

What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?


Answer:

Given: The amount deposited in bank is rupees 500.


The Compound interest is given by: A = P


Here p: principle, r: interest rate t: time in years, n: number of times interest is compounded in a year.


∴ At the end of first year A = 500 = 500(1.1)


∴ At the end of 10 years A = 500 = 500(1.1)10



Question 32.

If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.


Answer:

Given: A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively.


Let a and b be the roots of the quadratic equation.


Here,


A.M = = 8


⇒ a+b = 16 —1


G.M = √ab = 5


⇒ (√ab)2 = 52


⇒ ab = 25 —2


The quadratic equation is given by,


x2x (Sum of roots) + (Product of roots) = 0


x2x (a + b) + (ab) = 0


x2 – 16x + 25 = 0 (From eq –1 and eq –2)


∴ The required quadratic equation is x2 – 16x + 25 = 0




Miscellaneous Exercise
Question 1.

Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.


Answer:

Let a and d be the first term and the common difference of the A.P. respectively.


It is known that the kth term of an A.P. is given by


ak = a + (k –1) d


∴ am + n = a + (m + n –1) d


am – n = a + (m – n –1) d


am = a + (m –1) d


Now,


L.H.S = am + n + am – n


= a + (m + n –1) d + a + (m – n –1) d


= 2a + (m + n –1 + m – n –1) d


= 2a + (2m – 2) d


= 2a + 2 (m – 1) d


=2 [a + (m – 1) d]


= 2am


= R.H.S


Thus, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.


Question 2.

If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.


Answer:

Let the three numbers in A.P. be a – d, a, and a + d.


According to question -


(a – d) + (a) + (a + d) = 24 … (1)


⇒ 3a = 24


∴ a = 8


and,

also product of these numbers is 440,

(a – d) a (a + d) = 440 … (2)


⇒ (8 – d) (8) (8 + d) = 440


⇒ (8 – d) (8 + d) = 55


⇒ 64 – d2 = 55


⇒ d2 = 64 – 55 = 9


⇒ d = 3


Therefore,


when d = 3, the numbers are 5, 8, and 11 and


when d = –3, the numbers are 11, 8, and 5.


Thus, the three numbers are 5, 8, and 11.


Question 3.

Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that

S3 = 3(S2 – S1)


Answer:

Let a and b be the first term and the common difference of the A.P. respectively.


Therefore,


S1 = (n/2)[2a + (n - 1)d] …(1)


S2 = (2n/2)[2a + (2n - 1)d] …(2)


S3 = (3n/2)[2a + (3n - 1)d] …(3)


From (1) and (2), we obtain


S2 - S1 = (2n/2)[2a + (2n - 1)d] - (n/2)[2a + (n - 1)d]


= (n/2)[{4a + (4n - 2)d} - {2a + (n - 1)d}]


= (n/2)[4a + 4nd - 2d - 2a - nd + d]


= (n/2)[2a + 3nd - d]


= (1/3) × (3n/2)[2a + (3n - 1)d]


= (1/3)S3


Thus, S3 = 3(S2 - S1)


Hence, the given result is proved.



Question 4.

Find the sum of all numbers between 200 and 400 which are divisible by 7.


Answer:

The numbers lying between 200 and 400 which are divisible by 7


are as follows: -


203, 210, 217, … 399


Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P.


∴First term, a = 203


Last term, l = 399


Common difference, d = 7


Let the number of terms of the A.P. be n.


∴ an = 399 = a + (n –1) d


⇒ 399 = 203 + (n –1) 7


⇒ 7 (n –1) = 196


⇒ n –1 = 28


⇒ n = 29


We know that -


Sum of n terms of an A.P(Sn) = (n/2)[a + l]


S29 = (29/2)[203 + 399]


= (29/2)[602]


= 29 × 301


= 8729


Thus, the required sum is 8729.



Question 5.

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.


Answer:

The integers from 1 to 100 which are divisible by 2 are as follows: -


2, 4, 6… 100


Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with both the first term and common difference equal to 2.


No. of terms of above A.P = 100/2 = 50


Last Term(l) = 100


Sum of n terms of an A.P(Sn) = (n/2)[a + l]


∴ S60 = (50/2)[2 + 100]


= 25 × 102


= 2550


The integers from 1 to 100 which are divisible by 5 are as follows: -


5, 10,… 100


Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with both the first term and common difference equal to 5.


No. of terms of above A.P = 100/5 = 20


Last Term(l) = 100


Sum of n terms of an A.P(Sn) = (n/2)[a + l]


∴ S20 = (20/2)[5 + 100]


= 10 × 105


= 1050


The integers which are divisible by both 2 and 5 are as follows: -


10, 20, … 100


Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with both the first term and common difference equal to 10.


No. of terms of above A.P = 100/10 = 10


Last Term(l) = 100


Sum of n terms of an A.P(Sn) = (n/2)[a + l]


∴ S10 = (10/2)[10 + 100]


= 5 × 110


= 550


∴Required sum = 2550 + 1050 – 550 = 3050


Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.



Question 6.

Find the sum of all two-digit numbers which when divided by 4, yields 1 as remainder.


Answer:

The two - digit numbers which when divided by 4, yield 1 as remainder, are as follows: -


13, 17, … 97


Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with first term 13 and common difference 4.


Last Term of the A.P(l) = 97


Let n be the number of terms of the A.P.


It is known that the nth term of an A.P. is given by -


an = a + (n –1) d


⇒ 97 = 13 + (n –1) (4)


⇒ 4 (n –1) = 84


⇒ n – 1 = 21


⇒ n = 22


Sum of n terms of an A.P. is given by -


Sn = (n/2)[a + l]


∴ S22 = (22/2)[13 + 97]


= 11 × 110


= 1210


Thus, the required sum is 1210.



Question 7.

If f is a function satisfying f (x + y) = f(x) f(y) for all such that

f(1) = 3 and , find the value of n.


Answer:

Given:

f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)

f (1) = 3


Taking x = y = 1 in (1), we obtain


f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Therefore, f(2) = 9

Taking x = 1 and y = 2 in (1), we obtain


f (1 + 2) = f (3) = f (1) f (2) = 3 × 9 = 27

Therefore, f(3) = 27

Similarly, Taking x = 1 and y = 3 in (1), we obtain


f (1 + 3) = f (4) = f (1) f (3) = 3 × 27 = 81

Therefore, f(4) = 81

∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.


We know that -


Sum of first n terms of G.P with first term 'a' and common ratio 'r' is given by -



It is known that,


This means that,
f(1) + f(2) + f(3) + .....f(n) = 120

Thus,







∴ n = 4


Thus, the value of n is 4.


Question 8.

The sum of some terms of G.P. is 316 whose first term and the common ratio are 6 and 2, respectively. Find the last term and the number of terms.


Answer:

Given: Sum of n terms of the G.P. be 316.
a = 6, r = 2


To Find: n and an

Formula used:

where Sn = Sum to n terms of G.P,
a = first term,
r = common ratio,
n = number of terms

It is given that the first term a is 6 and common ratio r is 2.

Applying the values in formula we get,

n = 6


nth term of G.P is given by -


an = a r n - 1


∴ Last term of the G.P = 6th term = ar6 - 1 = (6)(2) 5 = (6)(32) = 192


Thus, the last term of the G.P. is 192.


Question 9.

The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.


Answer:

Let a and r be the first term and the common ratio of the G.P. respectively.


∴ a1 = 1


nth term of G.P is given by -


an = arn - 1


a3 = ar2 = (1)r2


a6 = ar4 = (1)r4


According to question -


a3 + a6 = 90


⇒ r2 + r4 = 90


⇒ r4 + r2 – 90 = 0


Assume r2 = t


⇒ t2 + t – 90 = 0


⇒ t2 + 10t - 9t – 90 = 0


⇒ t(t + 10) - 9(t + 10) = 0


⇒ (t - 9)(t + 10) = 0


∴ t =9 or t = - 10(invalid because t = r2, so t can't be - ve.)


∴ r2 = 9 ( Taking real roots)


∴ r = 3


Thus, the common ratio of the G.P. is 3.



Question 10.

The sum of three numbers in G.P. is 66. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.


Answer:

Let the three numbers in G.P. be a, ar, and ar2.


According to question -


a + ar + ar2 = 66


⇒ a(1 + r + r2) = 66


…(1)


Given that -


(a – 1), (ar – 7), (ar2 – 21) forms an A.P.


thus the common difference between the consecutive terms will be equal.


∴ (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)


⇒ ar – a – 6 = ar2 – ar – 14


⇒ ar2 – 2ar + a = 8


⇒ a(r2 – 2r + 1) = 8


…(2)


Comparing equations (1) & (2), we get -



On Cross - multiplying,


⇒66(r2 – 2r + 1) = 8(1 + r + r2)


⇒7(r2 – 2r + 1) = (1 + r + r2)


⇒ 7r2 – 14r + 7 = 1 + r + r2


⇒ 6r2 – 16r + 6 = 0


⇒ 6r2 – 12r - 3r + 6 = 0


⇒ 6r (r – 2) – 3 (r – 2) = 0


⇒ (6r – 3) (r – 2) = 0


∴ r = 2 or r = 1/2


When r = 2, a = 8


When r = 1/2, a = 32


Therefore,


when r = 2, the three numbers in G.P. are 8, 16, and 32.


when r = 1/2, the three numbers in G.P. are 32, 16, and 8.


Thus, in either case, the three required numbers are 8, 16, and 32.


Question 11.

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.


Answer:

Let the terms of G.P. be T1, T2, T3, T4, … T2n.


Number of terms = 2n


According to question,


T1 + T2 + T3 + … + T2n = 5 [T1 + T3 + … + T2n–1]


⇒ T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0


⇒ T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1]


Let the G.P. be a, ar, ar2, ar3, …



⇒ ar = 4a


∴ r = 4


Thus, the common ratio of the G.P. is 4.


Question 12.

The sum of the first four terms of an A.P. is 66. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.


Answer:

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.


Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d


Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + (n – 1) d] = 4a + (4n – 10) d


According to question -


4a + 6d = 66


⇒ 4(11) + 6d = 66 [∵ a = 11 (given)]


⇒ 6d = 12


⇒ d = 2


and,


4a + (4n –10) d = 112


⇒ 4(11) + (4n – 10)2 = 112


⇒ (4n – 10)2 = 68


⇒ 4n – 10 = 34


⇒ 4n = 44


⇒ n = 11


Thus, the number of terms of the A.P. is 11.



Question 13.

If , then show that a, b, c and d are in G.P.


Answer:

It is given that,



On cross multiplying, we get -


(a + bx)(b - cx) = (b + cx)(a - bx)


⇒ ab - acx + b2x - bcx2 = ab - b2x + acx - bcx2


⇒ 2b2x = 2acx


⇒ b2 = ac


…(1)


Also,



On cross multiplying, we get -


⇒ (b + cx)(c - dx) = (c + dx)(b - cx)


⇒ bc - bdx + c2x - cdx2 = bc + bdx - c2x - cdx2


⇒ 2c2x = 2bdx


⇒ c2 = bd


…(2)


From (1) and (2), we obtain



Thus, a, b, c, and d are in G.P.



Question 14.

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P.
Prove that P2Rn = Sn


Answer:

Let the G.P. be a, ar, ar2, ar3, … arn - 1


According to question -




[∵ Sum of first n natural numbers is n(n + 1)/2]




Now,


L.H.S = P2Rn






= Sn


= R.H.S


Hence, P2Rn= Sn



Question 15.

The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that

(q – r )a + (r – p )b + (p – q )c = 0


Answer:

Let A = first term of the AP.
and
Let d = common difference of the AP


Let n be the number of terms of the A.P.


It is known that the nth term of an A.P. is given by -


an = a + (n –1) d


pth term of A.P. is given by -
a = A + (p - 1).d.......(1)


qth term of A.P. is given by -
b = A + (q - 1).d.......(2)


rth term of A.P. is given by -
c = A + (r - 1).d........(3)
Subtracting (2) from (1) , (3) from (2) and (1) from (3), we get
a - b = (p - q).d......(4)
b - c = (q - r).d........(6)
c - a = (r - p).d.......(6)



Multiply (4),(6) and (6) by c, a and b respectively, we have

c.(a - b) = c.(p - q).d......(4)
a.(b - c) = a.(q - r).d........(6)
b.(c - a) = b.(r - p).d.......(6)


Adding (4),(6) and (6), we get -


c.(a - b) + a.(b - c) + b.(c - a) = c.(p - q).d + a.(q - r).d + b.(r - p).d
ac - bc + ab - ac + bc - ab = a.(q - r).d + b.(r - p).d + c.(p - q).d


a.(q - r).d + b.(r - p).d + c.(p - q).d = 0


Now since d is common difference, it should be non zero
Hence
a(q - r) + b(r - p) + c(p - q) = 0


Question 16.

If are in A.P., prove that a, b, c are in A.P.


Answer:

Given that are in AP.


If are in AP


Adding 1 to each term


are in AP


are in AP


are in AP


Divide each term by


are in AP


Hence, a, b, c are in AP


Hence Proved.



Question 17.

If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.


Answer:

We know that a, ar, ar2, ar3,… are in G.P. with first term a & common ratio r.


Given a, b, c, d are in G.P.


So, a = a


b = ar


c = ar2


d = ar3


We want to show that


(an + bn), (bn + cn), (cn + dn) are in GP i.e to show common ratio are same



Now,


L.H.S


putting b = ar, c = ar2






R.H.S


putting c = ar2, d = ar3, b = ar









Thus, L.H.S = R.H.S


Hence, (an + bn), (bn + cn), (cn + dn) are in GP.



Question 18.

If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p): (q – p) = 17:16.


Answer:

Given that a and b are roots of x2−3x + p=0


∴ a + b = 3 and ab= p …(1)


[∵ If α and β are roots of the equation ax2 + bx + c=0 then α + β=−b/a and αβ=c/a.]


It is given that c and d are roots of x2−12x + q=0


∴ c + d = 12 and cd=q …(2)


[∵ If α and β are roots of the equation ax2 + bx + c=0 then α + β=−b/a and αβ=c/a.]


Also given that a, b, c, d are in G.P.


Let a, b, c, d be the first four terms of a G.P.


So, a = a


b = ar


c = ar2


d = ar3


Now,


L.H.S






Now, From (1)


a + b=3


⇒ a + ar=3


⇒ a(1 + r)=3 ...(3)


From (2),


c + d=12


⇒ ar2 + ar3=12


⇒ ar2(1 + r)=12 ...(4)


Dividing equation (4) by (3), we get -


r2 = 4


∴ r4 = 16


putting the value of r4 in L.H.S, we get -




Hence proved.


Question 19.

The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that



Answer:

Here, the two numbers are a and b.


Arithmetic Mean = AM = (a + b)/2


& Geometric Mean = GM = √(ab)


According to question -




Applying component dividend







Applying Component Dividend





Squaring both sides






Hence, Proved.



Question 20.

If a, b, c are in A.P.; b, c, d are in G.P. and

are in A.P. prove that a, c, e are in G.P.


Answer:

It is given that a, b, c are in AP


∴ b = (a + c)/2 …(1)


Also given that b, c, d are in GP


∴ c2 = bd …(2)


Also,


are in AP


So, their common difference is same








…(3)


We need to show that a, c, e are in GP


i.e c2 = ae


From (2), we have


c2 = bd


Putting value of






⇒ c(c + e) = e(a + c)


⇒ c2 + ce = ea + ec


⇒ c2 = ea


Thus, a, c, e are in GP.


Hence, Proved.



Question 21.

Find the sum of the following series up to n terms:

6 + 66 + 666 + …


Answer:

The given sum is not in GP but we can write it as follows: -


Sum = 6 + 66 + 666 + …to n terms


= 6(1) + 6(11) + 6(111) + …to n terms


taking 6 common


= 6[1 + 11 + 111 + …to n terms]


divide & multiply by 9


= (6/9)[9(1 + 11 + 111 + …to n terms)]


= (6/9)[9 + 99 + 999 + …to n terms]


= (6/9)[(10 - 1) + (100 - 1) + (1000 - 1) + …to n terms]


= (6/9)[(10 - 1) + (102 - 1) + (103 - 1) + …to n terms]


= (6/9)[{10 + 102 + 103 + …n terms} - {1 + 1 + 1 + …n terms}]


= (6/9)[{10 + 102 + 103 + …n terms} - n]


Since 10 + 102 + 103 + …n terms is in GP with


first term(a) = 10


common ratio(r) = 102/10 = 10


We know that


Sum of n terms = (As r>1)


putting value of a & r


10 + 102 + 103 + …n terms




Hence, Sum



Question 22.

Find the sum of the following series up to n terms:

6 + . 66 + . 666 + …


Answer:

The given sum is not in GP but we can write it as follows: -


Sum = .6 + .66 + .666 + …to n terms


= 6(0.1) + 6(0.11) + 6(0.111) + …to n terms


taking 6 common


= 6[0.1 + 0.11 + 0.111 + …to n terms]


divide & multiply by 9


= (6/9)[9(0.1 + 0.11 + 0.111 + …to n terms)]


= (6/9)[0.9 + 0.99 + 0.999 + …to n terms]






Since is in GP with


first term(a) = 1/10


common ratio(r) = 10 - 2/10 - 1 = 10 - 1 = 1/10


We know that


Sum of n terms = (As r<1)


putting value of a & r






Hence, Sum



Question 23.

Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.


Answer:

The given series is in the form of multiplication of two different APs.


So, the nth term of given series is equal to the multiplication of their nth term.


The First AP is given as follows: -


2, 4, 6…


where, first term(a) = 2


common difference(d) = 4 - 2 = 2



nth term = a + (n - 1)d


= 2 + (n - 1)2


= 2 + 2n - 2


= 2n


The Second AP is given as follows: -


4, 6, 8…


where, first term(a) = 4


common difference(d) = 6 - 4 = 2



nth term = a + (n - 1)d


= 4 + (n - 1)2


= 4 + 2n - 2


= 2n + 2


Now,


an = [nth term of 2, 4, 6…] × [nth term of 4, 6, 8…]


= (2n) × (2n + 2)


= 4n2 + 4n


Thus, the nth term of series 2 × 4 + 4 × 6 + 6 × 8 + ... is


an = 4n2 + 4n


∴ a20 = 4 × (20)2 + 4 × 20 = 1600 + 80 = 1680


Hence, 20th term of series is 1680.



Question 24.

Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …


Answer:

This given series is neither an AP nor GP.


Let


Sn = 3 + 7 + 13 + 21 + 31 + … + an - 1 + an …(1)


Sn = 0 + 3 + 7 + 13 + 21 + 31 + … + an - 2 + an - 1 + an …(2)


Subtracting (2) from (1)


Sn - Sn = (3 - 0) + [(7 - 3) + (13 - 7) + (21 - 13) + … + (an - 1 - an - 2)


+ (an - an - 1) - an


⇒ 0 = 3 + [4 + 6 + 8 + … an - 1] - an


⇒ an = 3 + [4 + 6 + 8 + … an - 1] …(3)


Now, 4 + 6 + 8 + … an - 1 is an AP.


Where,


first term(a) = 4


common difference(d) = 6 - 4 = 2


We know that,


Sum of n terms of AP = (n/2)[2a + (n - 1)d]


putting n = n - 1, a = 4, d =2


[4 + 6 + 8 + … to (n - 1) terms] = (n - 1)/2 × [2a + (n - 1 - 1)d]


= (n - 1)/2 × [2(4) + (n - 2)2]


= (n - 1)/2 × [8 + 2n - 4]


= (n - 1)/2 × [2n + 4]


= (n - 1)/2 × 2(n + 2)


= (n - 1)(n + 2)



an = 3 + [4 + 6 + 8 + … an - 1]


= 3 + (n - 1)(n + 2)


= 3 + n2 + 2n - n - 2


= 3 + n2 + n - 2


= n2 + n + 1


Now,


Sn











Thus, the required sum is .



Question 25.

If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that


Answer:

According to question -


S1 = 1 + 2 + 3 + … + n =


S2 = 12 + 22 + 32 + … + n2 =


S3 = 13 + 23 + 33 + … + n3 =


Now,


R.H.S = S3(1 + 8S1)










= R.H.S


Hence, L.H.S = R.H.S


Hence Proved.



Question 26.

Find the sum of the following series up to n terms:



Answer:

The nth term of series is



Now, first solve the numerator & denominator separately


13 + 23 + 33 + … + n3 …(1)


Also,


1 + 3 + 5 + … + n terms


This is an AP.


whose first term(a) =1 & common difference(d) = 3 - 1 = 2


Now, sum of n terms of AP is


Sn = (n/2)[2a + (n - 1)d]


= (n/2)[2(1) + (n - 1)2]


= (n/2)[2 + 2n - 2]


= (n/2)[2n]


= n2


∴ Sn = n2 …(2)


Now,



putting values from (1) & (2)






Now, Finding Sum of n terms of Series













Thus, the required sum is



Question 27.

Show that



Answer:

Taking L.H.S



first we will solve the numerator & denominator separately


Let numerator be


S1 = 1 × 22 + 2 × 32 + 3 × 42 + … + n × (n + 1)2


nth term is n × (n + 1)2


Let an = n(n + 1)2


= n(n2 + 2n + 1)


= n3 + 2n2 + n


Now, S1











Let denominator be


S2 = 12 × 2 + 22 × 3 + … + n2 × (n + 1)


nth term is n2 × (n + 1)


Let bn = n2(n + 1) = n3 + n2


Now,


S2









Now,


L.H.S





= R.H.S


Hence, L.H.S = R.H.S


Hence Proved.



Question 28.

A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?


Answer:

Amount Paid to buy tractor = Rs. 12,000


Farmer Pays Cash = Rs. 6000


Remaining Balance = 12000 - 6000 = 6000


Annual Instalment = Rs 500 + [email protected]% on unpaid amount


1st Instalment


Unpaid Amount = Rs. 6000


Interest on Unpaid Amount = (12/100) × 6000 = 720


Amount of Instalment = Rs. 500 + Rs. 720 = Rs. 1220


2nd Instalment


Unpaid Amount = Rs. (6000 - 500) = Rs. 5500


Interest on Unpaid Amount = (12/100) × 5500 = 6600


Amount of Instalment = Rs. 500 + Rs. 660 = Rs. 1160


3rd Instalment


Unpaid Amount = Rs. (5500 - 500) = Rs. 5000


Interest on Unpaid Amount = (12/100) × 5000 = 600


Amount of Instalment = Rs. 500 + Rs. 600 = Rs. 1100


Total no. of Instalments = 6000/500 = 12


Thus, Annual Instalments are 1220, 1160, 1100, …upto 12 terms


Since the common difference between the consecutive terms is constant. Thus, Annual Instalments are in AP.


Here


first term(a) = 1220


Common difference(d) = 1160 - 1220 = - 60


Number of terms(n) = 12


Total amount paid in 12 instalments is given by -


Sn = (n/2)[2a + (n - 1)d]


∴ S12 = (12/2)[2(1220) + (12 - 1)( - 60)]


= 6[2440 + 11( - 60)]


= 6[2440 - 660]


= 6 × 1780


= 10680


Hence, total amount paid in 12 Instalments = Rs 10680


Hence,


Total Cost of Tractor


= Amount paid earlier + Amount paid in 12 Instalments


= Rs. (6000 + 10680)


= Rs. 16680



Question 29.

Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?


Answer:

Amount Paid to buy scooter = Rs. 22,000


Shamshad Pays Cash = Rs. 4000


Remaining Balance = Rs. (22000 - 4000) = 18000


Annual Instalment = Rs 1000 + [email protected]% on unpaid amount


1st Instalment


Unpaid Amount = Rs. 18000


Interest on Unpaid Amount = (10/100) × 18000 = 1800


Amount of Instalment = Rs. 1000 + Rs. 1800 = Rs. 2800


2nd Instalment


Unpaid Amount = Rs. (18000 - 1000) = Rs. 17000


Interest on Unpaid Amount = (10/100) × 17000 = 1700


Amount of Instalment = Rs. 1000 + Rs. 1700 = Rs. 2700


3rd Instalment


Unpaid Amount = Rs. (17000 - 1000) = Rs. 16000


Interest on Unpaid Amount = (10/100) × 16000 = 1600


Amount of Instalment = Rs. 1000 + Rs. 1600 = Rs. 2600


Total no. of Instalments = 18000/1000 = 18


Thus, Annual Instalments are 2800, 2700, 2600, …upto 18 terms


Since the common difference between the consecutive terms is constant. Thus, Annual Instalments are in AP.


Here


first term(a) = 2800


Common difference(d) = 2700 - 2800 = - 100


Number of terms(n) = 18


Total amount paid in 12 instalments is given by -


Sn = (n/2)[2a + (n - 1)d]


∴ S18 = (18/2)[2(2800) + (18 - 1)( - 100)]


= 9[5600 + 17( - 100)]


= 9[5600 - 1700]


= 9 × 3900


= 35100


Hence, total amount paid in 12 Instalments = Rs 35100


Hence,


Total Cost of Tractor


= Amount paid earlier + Amount paid in 12 Instalments


= Rs. (4000 + 35100)


= Rs. 39100



Question 30.

A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letters is mailed.


Answer:

According to question -


No. of letters in 1st set = 4


No. of letters in 2nd set = 4 × 4 = 16


No. of letters in 3rd set = 16 × 4 = 64


Hence, the sequence is


i.e. 4, 16, 64, …


This is a GP as


(16/4) = 4 & (64/16) = 4


Common ratio(r) = 4


First Term(a) = 4


Total no. of letters mailed upto 8th set is given by putting values of a, r & n(=8) in


(as r>1)





= 4 × 21845


= 87380


Hence, total number of letters posted after 8th set is 87380.


Given that -


Postage Charge per Letter = 50 paise


Hence, the amount spend on the postage of 8738 letter is


= Rs. [(50/100) × 87380]


= Rs. 43690



Question 31.

A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.


Answer:

In simple interest, the interest remains same in all year.


Interest per year = 10000 × 5% = 500


Hence,


Amount in 1st year = Rs. 10000


Amount in 2nd year = Amount in 1st year + Interest


= 10000 + 500


= 10500


Amount in 3rd year = Amount in 2nd year + Interest


= 10500 + 500


= 11000


Hence, the series becomes


10000, 10500, 11000,…


Since the common difference between the consecutive terms is constant. Thus, Above Series are in AP.


Where,


first term(a) = 10000


common difference(d) = 10500 - 10000 = 500


Amount in 15th year is given by putting n = 15 in


an = a + (n - 1)d


∴ a15 = 10000 + (15 - 1)500


= 10000 + 14(500)


= 10000 + 7000


= 17000


Thus, amount in 15th year is Rs. 17000.


Also, Amount after 20 years


= a21


= 10000 + (21 - 1)500


= 10000 + 20 × 500


= 10000 + 10000


= 20000


Thus, amount after 20th year is Rs. 20000.



Question 32.

A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.


Answer:

By Using Formula


A = P[1 - (r/100)]n


Here, P = principal = Rs. 15625


r = rate of depreciation = 20%


n = number of years = 5 years


A be the depreciated value


Putting values in the formula,


Depreciated Value = 15625[1 - (20/100)]5


= 15625[1 - (1/5)]5


= 15625(4/5)5


= (15625 × 1024)/3125


= 5 × 1024


= 5120


Thus, the depreciated value of the machine after 5 years is


Rs. 5125.



Question 33.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.


Answer:

Let total work = 1


and let total work be completed in 'n' days


Work done in 1 day = 1/n


This is the work done by 150 workers


Work done by 1 worker in one day = 1/150n


Case 1: -


No. of workers = 150


Work done per worker in 1 day = 1/150n


Total work done in 1 day = 150/150n


Case 2: -


No. of workers = 146


Work done per worker in 1 day = 1/150n


Total work done in 1 day = 146/150n


Case 3: -


No. of workers = 142


Work done per worker in 1 day = 1/150n


Total work done in 1 day = 142/150n


Given that


In this manner it took 8 more days to finish the work i.e. work finished in (n + 8) days.




…(1)


Now,


is an AP


where,


first term(a) = 150


common difference(d) = 146 - 150 = - 4


we know that


Sum of n terms of AP(Sn) = (n/2)[2a + (n - 1)d]


putting n = n + 8, a = 150 & d = - 4


Sn + 8 = [(n + 8)/2] × [2(150) + (n + 8 - 1)( - 4)]


= [(n + 8)/2] × [300 + (n + 7)( - 4)]


= [(n + 8)/2] × [300 - 4n - 28]


= [(n + 8)/2] × [272 - 4n]


= (n + 8) × (136 - 2n)


= - 2n2 + 120n + 1088


From (1),


Sn + 8 = 150n


⇒ - 2n2 + 120n + 1088 = 150n


⇒ - 2n2 - 30n + 1088 = 0


⇒ - 2(n2 + 15n - 544) = 0


⇒ (n2 + 15n - 544) = 0


⇒ n2 + 32n - 17n - 544 = 0


⇒ n(n + 32) - 17(n + 32) = 0


⇒ (n - 17)(n + 32) = 0


∴ n = 17


because n = - 32 is invalid as no. of days can't be - ve.


Hence, n =17


Thus, the work was completed in n + 8 days i.e. 17 + 8 = 25 days