Sequences And Series

The n^th term of the sequence is given as a_n = 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

The given n^th 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

The nth term of the sequence is a_n = 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

The n^th 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

The n^th term of the sequence is given as a_n= (-1)^n-1 5^{n + 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 5^{1 + 1} = (-1)⁰ 5² = 25 (∵ a⁰ =1)

a₂ = (-1)^2-1 5^{2 + 1} = (-1)¹ 5³ = -125

a₃ = (-1) ^3-1 5^{3 + 1} = (-1)² 5⁴ = 625

a₄ = (-1)^4-1 5^{4 + 1} = (-1)³ 5⁵ = -3125

a₅= (-1)^5-1 5^{5 + 1} = (-1)⁴ 5⁶ = 15625

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

The n^th 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

The n^th term of the sequence is

a_n = 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

The given n^th term of the sequence is

………….1

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

The given n^th term of the sequence is a_n= (-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

In the given expression the value of n^th term is

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

Given here are a₁ = 3,

a_n = 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 (∵ a_n-1 = a_3-1 = 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 + …….

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

Here given, a₁ = a₂ = 2, a_n = a_n-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) + ………

Here given a1 = a2 =1

And a_n = a_n-1 + a_n-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

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. =

⇒ S_n = 1002001

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

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. =

⇒ S_n = 179 × 550 = 98,450.

Given, first term a = 2

Let d be the common difference

Let S_n denote sum of n terms,

t_n denote n^th term.

t_n = a + (n – 1)d

S₁ = Sum of first five terms.

S₂ = Sum of next five terms.

S₁ = 10 × (1 + d)

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

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

S_n = 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

Let t_n = n^th tern of AP

t_n = a + (n – 1)d

S_n = sum of n terms of AP

Let a be the first term and d common difference.

Given, in an AP

p^th term = 1/q

………………………………..(I)

q^th 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 .

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

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

S_n = 116

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

Putting the value of a, d and S_n

⇒ 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

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

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

k^th term = a_k = 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

∴

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

We know,

∴

Comparing the coefficients of n² on both sides, we obtain

⇒ d = 2q

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

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.

⇒
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 18^th term of both the A.P.s is 179: 321

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 =

S_p = S_q

⇒ 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

⇒

∴

⇒

⇒ S_p+q = 0

Given:S_p = a, S_q = b, and S_r = c

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.

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

Given,

Sum of m terms of A.P. = S_m

Sum of n terms of A.P. = S_n

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

⇒

∴ Ratio of m^th and n^th term is (2m – 1) : (2n – 1).

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

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

m^th term of A.P. = t_m = 164 ……………………….(III)

⇒ t_m = 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

m^th term = t_m = 164 = 8 + (m – 1)6

⇒ 164 – 8 = (m – 1)6

⇒ m – 1 = 156/6

⇒ m – 1 = 26

⇒ m = 26 + 1

∴ m = 27

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.

A.M. of a and b =

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

∴

⇒ (a + b) (a^n-1+ b^n-1) = 2aⁿ + 2bⁿ

⇒ aⁿ + ab^n-1 + ba^n-1 + bⁿ = 2aⁿ + 2bⁿ

⇒ ab^n-1 + ba^n-1 = aⁿ + bⁿ

⇒ ab^n-1 – bⁿ = aⁿ – ba^n-1

⇒ b^n-1 (a – b) = a^n-1 (a – b)

⇒

⇒

⇒ n – 1 = 0

∴ n = 1

Let A₁, A₂, … A_m be m numbers such that 1, A₁, A₂, … A_m, 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.

A_m-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

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

n^th term of A.P. = t_n = a + (n – 1)d

30^th term = t₃₀ = 100 + (30 – 1) × 5

= 100 + 29 × 5 = 100 + 145

t₃₀ = 245

Man will pay Rs 245 in 30^th instalment.

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)

S_n = 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

Given: G.P.

We know that in G.P a_n = ar^n-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, n^th team of the G.P :

∴ n^th term of the given G.P. is

20^th term of the given G.P. is

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

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

That is,

a₈ = 192 and r = 2.

We know that in G.P a_n = ar^n-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

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

We know that in G.P a_n = ar^n-1

Here, n: number of terms

a: First term

r: common ratio

Here,

a₅ = ar^5-1 = ar⁴

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

Similarly,

a₈ = ar^8-1 = ar⁷

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

a₁₁ = ar^11-1 = ar¹⁰

⇒ s = ar¹⁰ (∵ 11^th 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

Given: a₄ = (a₂)² and a = –3

We know that in G.P a_n = ar^n-1

∴ a₂ = (–3) × r^2-1 = –3r(∵ a = –3)

Similarly,

a₄ = –3r³

∴ –3r³ = (-3r)² (∵ a₄ = (a₂)² )

∴ –3r³ = 9r²

⇒ r = –3

∴ r = –3

Now,

a₇ = ar^7-1

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

⇒ a₇ = ( –3)⁷ = – 2187.

∴ a₇ = – 2187

A)

Given: 2, 2√2, 4….. and a_n = 128

Here, in the above G.P.

a = 2

Common ratio(r) =

We know that in G.P a_n = ar^n-1

∴ a_n = 2 × (√2)^n-1

⇒ 128 = 2 × (√2)^n-1

⇒ (√2)^n-1 = 128/2 = 64

⇒ = 64 (∵ √x = x^1/2)

Apply ln on both sides

We get

⇒

⇒

⇒

⇒ n – 1 = 12

⇒ n = 13

∴ a₁₃ = 128

B)

Given: √3, 3, 3√3 ….. and a_n = 729

Here, in the above G.P.

a = √3

Common ratio(r)

We know that in G.P a_n = ar^n-1

∴ a_n = √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 a_n =

Here, in the above G.P.

a =

Common ratio(r) = =

We know that in G.P a_n = ar^n-1

∴ a_n =

⇒

Apply ln on both sides

We get

⇒

⇒

⇒ n = 9

∴ a₉ =

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 .

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]

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

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]

Given: x³, x^5, 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:

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)

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: 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

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

Given: a = 729 and a₇ = 64.

nth term of a G.P is given by ar^n-1

∴ a₇ = ar^7-1

⇒ 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

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 ar^n-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,…….

Given: 4^th, 10^th and 16^th 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 ar^n-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

Given: sequence: 8, 88, 888, 8888… .

S_n = 8 + 88 + 888 + 8888 + …….

⇒ S_n = 8[1 + 11+ 111 + 1111 +…….]

⇒ S_n = × [9 + 99+ 999 + 9999 +…….]

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

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

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

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

⇒S_n =

⇒S_n =

⇒S_n =

∴

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.

Given: The sequences a, ar, ar²,…ar^{n – 1} and A, AR, AR², … AR^{n – 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

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

Given p^th, q^th and r^thterms of a G.P are a, b and c, respectively

Here

a_p = a = ar^p-1

a_q = b = ar^q-1

a_r = c = ar^r-1

Now,

a^{q – r} × b^{r – p} × c^{P – q} = (ar^{p - 1})^{q - r} × (ar^{q - 1})^{r - p} × (ar^r-1)^{p – q}

⇒ a^{q – r} × b^{r – p} × c^{P – 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)})

⇒ a^{q – r} × b^{r – p} × c^{P – 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)})

⇒ a^{q – r} × b^{r – p} × c^{P – q} = (a^{(q – r + r – p + p - q)} × r^{(pq – q – pr +r + qr - r –pq + p +pr – p –qr + q)})

⇒ a^{q – r} × b^{r – p} × c^{P – q} = (a⁰ × r⁰) = 1

∴ a^{q – r} × b^{r – p} × c^{P – q} = 1

Hence proved.

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

a_n = b = ar^n-1 —1

Here,

P = Product of n terms

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

⇒ P = (a × a × …a) × (r × r² × …r^n-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

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,

a_n+1 = ar^n+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 = .

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

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.

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

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

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

(a – b)² = (a + b)² – 4ab, we get

(a – b)² = 4A² – 4G² = 4 (A²–G²)

(a – b)² = 4 (A + G) (A – G)

(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 .

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 2^nd hour = ar² = 30 × 2² = 120

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

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

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

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

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,

x²– x (Sum of roots) + (Product of roots) = 0

x² – x (a + b) + (ab) = 0

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

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

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

a_k = a + (k –1) d

∴ a_{m + n} = a + (m + n –1) d

a_{m – n} = a + (m – n –1) d

a_m = a + (m –1) d

Now,

L.H.S = a_{m + n} + a_{m – 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]

= 2a_m

= 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.

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,

(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.

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.

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

S₂₉ = (29/2)[203 + 399]

= (29/2)[602]

= 29 × 301

= 8729

Thus, the required sum is 8729.

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(S_n) = (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(S_n) = (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(S_n) = (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.

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 -

a_n = 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 -

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

∴ S₂₂ = (22/2)[13 + 97]

= 11 × 110

= 1210

Thus, the required sum is 1210.

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

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

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

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

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

Thus,

∴ n = 4

Thus, the value of n is 4.

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

To Find: n and a_n

Formula used:

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

∴ n = 6

nth term of G.P is given by -

a_n = a r ^{n - 1}

∴ Last term of the G.P = 6th term = ar^{6 - 1} = (6)(2) ⁵ = (6)(32) = 192

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

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 -

a_n = ar^{n - 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.

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.

Let the terms of G.P. be T₁, T₂, T₃, T₄, … T_2n.

Number of terms = 2n

According to question,

T₁ + T₂ + T₃ + … + T_2n = 5 [T₁ + T₃ + … + T_2n–1]

⇒ T₁ + T₂ + T₃ + … + T_2n – 5 [T₁ + T₃ + … + T_2n–1] = 0

⇒ T₂ + T₄ + … + T_2n = 4 [T₁ + T₃ + … + T_2n–1]

Let the G.P. be a, ar, ar², ar³, …

⇒ ar = 4a

∴ r = 4

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

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.

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.

Let the G.P. be a, ar, ar², ar³, … ar^{n - 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

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 -

a_n = a + (n –1) d

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

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

r^th 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

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.

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.

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.

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.

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.

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

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

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

∴

n^th 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

∴

n^th term = a + (n - 1)d

= 4 + (n - 1)2

= 4 + 2n - 2

= 2n + 2

Now,

a_n = [n^th term of 2, 4, 6…] × [n^th term of 4, 6, 8…]

= (2n) × (2n + 2)

= 4n² + 4n

Thus, the n^th term of series 2 × 4 + 4 × 6 + 6 × 8 + ... is

a_n = 4n² + 4n

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

Hence, 20^th term of series is 1680.

This given series is neither an AP nor GP.

Let

S_n = 3 + 7 + 13 + 21 + 31 + … + a_{n - 1} + a_n …(1)

S_n = 0 + 3 + 7 + 13 + 21 + 31 + … + a_{n - 2} + a_{n - 1} + a_n …(2)

Subtracting (2) from (1)

S_n - S_n = (3 - 0) + [(7 - 3) + (13 - 7) + (21 - 13) + … + (a_{n - 1} - a_{n - 2})

+ (a_n - a_{n - 1}) - a_n

⇒ 0 = 3 + [4 + 6 + 8 + … a_{n - 1}] - a_n

⇒ a_n = 3 + [4 + 6 + 8 + … a_{n - 1}] …(3)

Now, 4 + 6 + 8 + … a_{n - 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)

∴

a_n = 3 + [4 + 6 + 8 + … a_{n - 1}]

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

= 3 + n² + 2n - n - 2

= 3 + n² + n - 2

= n² + n + 1

Now,

S_n

Thus, the required sum is .

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.

The n^th 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

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

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

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

= (n/2)[2n]

= n²

∴ S_n = n² …(2)

Now,

putting values from (1) & (2)

Now, Finding Sum of n terms of Series

Thus, the required sum is

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

n^th term is n × (n + 1)²

Let a_n = n(n + 1)²

= n(n² + 2n + 1)

= n³ + 2n² + n

Now, S₁

Let denominator be

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

n^th term is n² × (n + 1)

Let b_n = n²(n + 1) = n³ + n²

Now,

S₂

Now,

L.H.S

= R.H.S

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

Hence Proved.

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 -

S_n = (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

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 -

S_n = (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

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 8^th 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

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

a_n = 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.

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.

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

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

S_{n + 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),

S_{n + 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