Arithmetic Progression

To Find: First four terms of given series.

(i) Given: n^th term of series is (5n + 2)

Put n=1, 2, 3, 4 in n^th term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term

a₁ = (51 + 2) = 7

a₂ = (52 + 2) = 12

a₃ = (53 + 2) = 17

a₄ = (54 + 2) = 22

First four terms of given series is 7, 12,17,22

ALTER: When you find or you have first term (a or a₁) and second term (a₂) then find the difference (a₂ - a₁)

Now add this difference in last term to get the next term

For example a₁= 7 and a₂= 12, so difference is 12 - 5 = 7

Now a₃ = 12 + 5 = 17, a₄ = 17 + 5 = 22

(This method is only for A.P)

NOTE: When you have nth term in the form of (an + b)

Then common difference of this series is equal to a.

This type of series is called A.P (Arithmetic Progression)

(Where a, b are constant, and n is number of terms)

(ii) Given: n^th term of series is

Put n=1, 2, 3, 4 in n^th term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term.

a₁ = =

a₂ = =

a₃ = =

a₄ = =

First four terms of given series are , , ,

(iii) Given: n^th term of series is (–1)^n–1 × 2^{n + 1}

Put n=1, 2, 3, 4 in n^th term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term.

a₁ = (–1)^1–1 × 2^{1 + 1} = (–1)⁰ × 2² = 14 = 4

a₂ = (–1)^2–1 × 2^{2 + 1} = (–1)¹ × 2³ = (–1)8 = (–8)

a₃ = (–1)^3–1 × 2^{3 + 1} = (–1)² × 2⁴ = 116 = 16

a₄ = (–1)^4–1 × 2^{4 + 1} = (–1)³ × 2⁵ = (–1)32 = (–32)

First four terms of given series are 4, –8 , 16 ,–32

To Find: First five terms of a given sequence.

Condition: n ≥ 2

Given: a₁ = 1, a_n = a_n–1 + 3 for n ≥ 2

Put n= 2 in n^th term (i.e. a_n), we have

a₂ = a_2–1 + 3 = a₁ + 3 = 1 + 3 = 4 (as a₁ = 1)

Put n= 3 in n^th term (i.e. a_n), we have

a₃ = a_3–1 + 3 = a₂ + 3 = 4 + 3 = 7 (as a₂ = 4)

Put n= 4 in n^th term (i.e. a_n), we have

a₄ = a_4–1 + 3 = a₃ + 3 = 7 + 3 = 10 (as a₃ = 7)

Put n= 5 in n^th term (i.e. a_n), we have

a₅ = a_5–1 + 3 = a₄ + 3 = 10 + 3 = 13 (as a₂ = 10)

First five terms of a given sequence is 1, 4, 7, 10, 13

To Find: First five terms of a given sequence.

Condition: n ≥ 2

Given: a₁ = –1, a_n = for n ≥ 2

Put n= 2 in n^th term (i.e. a_n), we have

a₂ = (as a₁ = –1 )

Put n= 3 in n^th term (i.e. a_n), we have

a₃ = (as a₂ = )

Put n= 4 in n^th term (i.e. a_n), we have

a₄ = (as a₃ = )

Put n= 5 in n^th term (i.e. a_n), we have

a₅ = (as a₃ = )

First five terms of a given sequence are –1, , , ,

To Find: 23^rd term of the AP

Given: The series is 7, 5, 3, 1, –1, –3, …

a₁= 7, a₂= 5 and d= 3–5= –2

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

So put n =23 in above formula, we have

a₂₃= a₁ + (23 - 1)(–2) = 7– 44 = –37

So 23^rd term of AP is equal to –37.

To Find: 20^th term of the AP

Given: The series is √2, 3√2, 5√2, 7√2, ….

a₁=√2, a₂= 3√2 and d= 3√2–√2= 2√2

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a₂₀= a₁ + (20 - 1)( 2√2)= √2 + 38√2= 39√2

So 20^rd term of AP is equal to 39√2.

To Find: n^th term of the AP

Given: The series is 8, 3, –2, –7, –12, ….

a₁=8, a₂= 3 and d= 3–8= –5

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a_n= a₁ + (n - 1)( –5) = 8– (5n–5) = 8– 5n + 5 = 13– 5n

So the n^th term of AP is equal to 13– 5n

To Find: n^th term of the AP

Given: The series is 1, , , , …

a₁=1, a₂= and d= –1=

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a_n= a₁ + (n - 1)( ) = 1– () = =()

So the n^th term of AP is equal to ()

To Find: we need to find n when a_n = 379

Given: The series is 9, 14, 19,24, 29, …. and a_n=379

a₁=9, a₂= 14 and d=14–9 = 5

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a_n= 379 = a₁ + (n–1)5

379 – 9 = (n–1)5 [subtract 9 on both side]

370 = (n–1 � � �)5

74 = (n–1) [Divide both side by 5]

n = 75^th

The 75^th term of this AP is equal to 379.

To Find: we need to find n when a_n = 0

Given: The series is 64, 60, 56, 52, 48, … and a_n= 0

a₁=64, a₂= 60 and d=60–64 = –4

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a_n= 0 = a₁ + (n–1)(–4)

0– 64 = (n–1)(–4) [subtract 64 on both sides]

– 64 = (n–1)(–4)

64 = (n–1)4 [Divide both side by ‘–‘]

16 = (n–1) [Divide both side by 4]

n = 17^th [add 1 on both sides]

The 17^th term of this AP is equal to 0.

To Find: we need to find a number of terms in the given AP.

Given: The series is 11, 18, 25, 32, 39, …. 207

a₁=11, a₂= 18,d=18–11 = 7 and a_n=207

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n–1)d

a_n= 207 = a₁ + (n–1)(7)

207– 11 = (n–1)(7) [subtract 11 on both sides]

196 = (n–1)(7)

28 = (n–1) [Divide both side by 7]

n = 29 [add 1 on both sides]

So there are 29 terms in this AP.

To Find: we need to find number of terms in the given AP.

Given: The series is 1, 1, …., –16 .

a₁=1= , a₂= 1=, d=()–()= and a_n= –16 =

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a_n= = a₁ + (n–1)()

– = (n–1)() [subtract on both sides]

= (n–1)() [Multiply both side by ] or [Divide both side by ]

27 = (n–1) [add 1 on both sides]

n =28

So there are 28 terms in this AP.

To Find: –47 is a term of the AP or not.

Given: The series is 5, 2, –1, –4, –7, ….

a₁=5, a₂= 2, and d=2–5 = –3 (Let suppose an = –47)

NOTE: n is a natural number.

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a_n = –47 = a + (n - 1)d

–47 = 5 + (n - 1)(–3)

–47–5 = (n - 1)(–3) [subtract 5 on both sides]

52 = (n - 1)(3) [Divide both side by ‘–‘]

17.33 = (n - 1) [Divide both side by 3]

18.33 = n [add 1 on both sides]

As n is not come out to be a natural number, So –47 is not the term of this AP.

To Find: AP and its 30^th term (i.e. a₃₀=?)

Given: a₅=5 and a₁₃=–3

Formula Used: a_n = a + (n - 1)d

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

By using the above formula, we have

a₅ = 5 = a + (5 - 1)d , and a₁₃ = –3 = a + (13 - 1)d

a + 4d =5 and a + 12d = –3

on solving above 2 equation, we and a + 12d = –3get

a = 9 and d= (–1)

So a₃₀ = 9 + 29(–1) = –20

AP is (9,8,7,6,5,4……) and 30^th term = –20

To Find: First term and number of terms.

Given: a₂= , a₃₁ = , and an =

Formula Used: a_n = a + (n - 1)d

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

By using above formula, we have

a₂ = = a + d and a₃₁ = = a + (31 - 1)d

on solving both equation, we get

a = 8 and d = –0.25

Now an = = 8 + (n - 1)( –0.25)

On solving the above equation, we get

n= 59

So the First term is equal to 8 and the number of terms is equal to 59.

Prove that: 29^th term is double the 19^th term (i.e. a₂₉ = 2a₁₉)

Given: a₉= 0

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

So a₉= 0 a + (9 - 1)d = 0

a + 8d = 0

a = (–8d) ….equation (i)

Now a₂₉ = a + (29 - 1)d and a₁₉ = a + (19 - 1)d

a₂₉= a + 28d and a₁₉ = a + 18d ….equation (ii)

By using equation (i) in equation (ii), we have

a₂₉= –8d + 28d and a₁₉ = –8d + 18d

a₂₉= 20d and a₁₉ = 10d

So a₂₉= 2a₁₉

HENCE PROVED

To Find: First term (a) and common difference (d)

Given: a₄= 3a₁ and a₇ = 2a₃ + 1

(Where a=a₁ is first term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

a₄= 3a₁ a + 3d= 3a 3d = 2a ….equation (i) and

a₇ = 2a₃ + 1 a + 6d= 2(a + 2d) + 1 2d= a + 1 ….equation (ii)

on solving both equation (i) & (ii), we get

a= 3 and d= 2

So the first term is equal to 3, and the common difference is equal to 2.

Show that: 18^th term of the AP is zero.

Given: 7a₇= 11a₁₁

(Where a₇ is Seventh term, a₁₁ is Eleventh term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

7(a + 6d) = 11(a + 10d)

7a + 42d = 11a + 110d 68d = (–4a)

a + 17d= 0 ….equation (i)

Now a₁₈ = a + (18 - 1)d

So a + 17d = 0 [by using equation (i)]

HENCE PROVED

[NOTE: If n times the n^th term of AP is equal to m times the m^th term of same AP then its (m + n)^th term is equal to zero]

To Find 28^th term from the end of the AP.

Given: The AP is 6, 9, 12, 15, 18, …., 102

a₁ = 6, a₂ = 9, d = 9–6 = 3 and l = 102

Formula Used: nth term from the end = l– (n–1)d

(Where lis last term and d is common difference of given AP)

By using nth term from the end = l– (n–1)d formula

28th term from the end = 102– 27d 102– 273 = 21

So 28^th term from the end is equal to 21.

To Find : 28^th term from the end of the AP.

Given: The AP is 7, 2, –3, –8, –13, …., –113

a₁ = 7, a₂ = 2, d = 2–7 = –5 and l = –113

Formula Used: nth term from the end = l– (n–1)d

(Where lis last term and d is common difference of given AP)

By using nth term from the end = l– (n–1)d formula

16th term from the end = (–113)– 15d (–113)–15(–5) = –38

So 16^th term from the end is equal to –38.

To Find : 3 - digit numbers divisible by 7.

First 3 - digit number divisible by 7 is 105

Second 3 - digit number divisible by 7 is 112 and

Last 3 - digit number divisible by 7 is 994.

Given: The AP is 105, 112, 119,…………..,994

a₁ = 105, a₂ = 112, d = 112–105 = 7 and a_n = 994

(Where a=a₁ is First term, a₂ is Second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

994 = 105 + (n - 1)7

889 = (n - 1)7

127 = (n - 1)

n = 128

So, There are total of 128 three - digit number which is divisible by 7.

To Find : 2 - digit numbers divisible by 3.

First 2 - digit number divisible by 3 is 12

Second 2 - digit number divisible by 3 is 15 and

Last 2 - digit number divisible by is 99.

Given: The AP is 12, 15, 18,…………..,99

a₁ = 12, a₂ = 15, d = 15–12 = 3 and a_n = 99

(Where a=a₁ is First term, a₂ is Second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

99 = 12 + (n - 1)3

87 = (n - 1)3

29 = (n - 1)

n = 30

So, There are total of 30 two - digit number which is divisible by 3.

Show that: sec θ₁sec θ₂ + sec θ₂sec θ₃ + …. + sec θ_n–1sec θn = .

Given: Given AP is θ₁, θ₂, θ₃, …., θ_n

a= θ_1, a₂= θ₂ and d= θ₂ - θ₁= θ₃ - θ₂= θ₄ - θ₃=…………= θ_n - θ_{n - 1}

sec θ₁sec θ₂ + sec θ₂sec θ₃ + …. + sec θ_n–1sec θn = + ………… +

Multiply both side by sin d

sin d (sec θ₁sec θ₂ + sec θ₂sec θ₃ + …. + sec θ_n–1sec θn)= + ………… +

[NOTE: sin(x - y)=sinxcosy - cosxsiny, & secθcosθ=1]

By using above formula on R.H.S. , we get

R.H.S. = tanθ₂ - tanθ₁ + tanθ₃ - tanθ_{2 +} tanθ₄ - tanθ₃ ………….. + tanθ_n - tanθ_{n - 1}

R.H.S. = tanθ_n - tanθ₁ (All the remaining term cancle out)

sin d (sec θ₁sec θ₂ + sec θ₂sec θ₃ + …. + sec θ_n–1sec θn)= tanθ_n - tanθ₁ (Divide sin d on both sides), we get

sec θ₁sec θ₂ + sec θ₂sec θ₃ + …. + sec θ_n–1sec θn= .

HENCE PROVED

To Find:

Given:

(Where T_n is nth term and d is common difference of given AP)

Formula Used: T_n = a + (n - 1)d

= (cross multiply)

3a + 9d = 2a + 12d a = 3d …….equation (i)

Now = = =

=

So =

To Find: The three numbers which are in AP.

Given: Sum and product of three numbers are 27 and 648 respectively.

Let required number be (a - d), (a), (a + d). Then,

(a - d) + a + (a + d) = 27 3a = 27 a = 9

Thus, the numbers are (9 - d), 9 and (9 + d).

But their product is 648.

(9 - d) × 9(9 + d)= 648

(9 - d)(9 + d)= 72

81 – d² = 72 d² = 9 d =3

When d=3 numbers are 6, 9, 12

When d= (3) numbers are 12, 9, 6

So, Numbers are 6, 9, 12 or 12, 9, 6.

To Find: The three numbers which are in AP.

Given: Sum and sum of the squares of three numbers are 21 and 165 respectively.

Let required number be (a - d), (a), (a + d). Then,

(a - d) + a + (a + d) = 21 3a = 21 a = 7

Thus, the numbers are (7 - d), 7 and (7 + d).

But their sum of the squares of three numbers is 165.

(7 - d)²7²(7 + d)²= 165

49 + d²14d + 49 + d² + 14d = 116

2d² = 18 d² = 9 d = 3

When d=3 numbers are 4, 7, 10

When d= (3) numbers are 10, 7, 4

So,Numbers are 4, 7, 10 or 10, 7, 4.

To Find: The angles of a quadrilateral.

Given: Angles of a quadrilateral are in AP with common difference = 10°.

Let the required angles be a, (a + 10°), (a + 20°) and (a + 30°).

Then, a + (a + 10°) + (a + 20°) + (a + 30°)=360° 4a + 60°= 360° a = 75°

NOTE: Sum of angles of quadrilateral is equal to 360°

So Angles of a quadrilateral are 75°, 85°, 95° and 105°.

To Find: The number

Given: The digits of a 3 - digit number are in AP, and their sum is 15.

Let required digit of 3 - digit number be (a - d), (a), (a + d). Then,

(a - d) + (a) + (a + d)=15 3a = 15 a = 5

(Figure show 3 digit number original number)

(Figure show 3 digit number in reversing form)

So, (5 + d)100 + 510 + (5 - d)1 = {(5d)100 + 510 + (5 + d)1} – 594

200d – 2d = – 594 d = –3 and a = 5

So the original number is 852

To Find: The number of terms common to both AP

Given: The 2 AP’s are 5, 9, 13, 17, …., 217 and 3, 9, 15, 21, …., 321

As we find that first common term of both AP is 9 and the second common term of both AP is 21

Let suppose the new AP whose first term is 9, the second term is 21, and the common difference is 21 – 9 = 12

NOTE: As first AP the last term is 217 and second AP last term is 321. So last term of supposing AP should be less than or equal to 217 because after that there are no common terms

Formula Used: T_n = a + (n - 1)d

(Where T_n is nth term and d is common difference of given AP)

217 a + (n - 1)d 9 + (n - 1)12 217

(n - 1)12 208 (n - 1) 17.33 n 18.33

So, Number of terms common to both AP is 18.

Show that: the sum of the interior angles of polygons with 3, 4, 5, 6, …. sides form an arithmetic progression.

To Find: The sum of the interior angles for a 21 - sided polygon.

Given: That the sum of the interior angles of a triangle is 180°.

NOTE: We know that sum of interior angles of a polygon of side n is (n – 2) x 180°.

Let a_n= (n – 2) x 180° Since a_n is linear in n. So it forms AP with 3, 4, 5, 6,……sides

{a_n is the sum of interior angles of a polygon of side n}

By using the above formula, we have

a_{21 =} (21 – 2) x 180°

a₂₁ =3420°

So, the Sum of the interior angles for a 21 - sided polygon is equal to 3420°.

To Find: The perimeter of the sixth inscribed equilateral triangle.

Given: Side of an equilateral triangle is 24 cm long.

As 2^nd triangle is formed by joining the midpoints of the sides of the first triangle whose side is equal to 24cm

[As shown in the figure]

So Side of a 2^nd equilateral triangle is 12 cm long [half of the first triangle side]

Side of 2^nd equilateral triangle = half of side of a 1^st equilateral triangle

Side of 3^rd equilateral triangle = half of side of a 2^nd equilateral triangle

…………. and So on

Therefore, Side of 6^th equilateral triangle = half of side of a 5^th equilateral triangle

So, Perimeter of a 6^th equilateral triangle is 3 times the side of a 6^th equilateral triangle

[NOTE: Perimeter of the triangle is equal to the sum of all three sides of the triangle, and in case of an equilateral triangle all sides are equal]

So,Perimeter of 6^th equilateral triangle = 30.75 = 2.25 cm

To Find: what amount will he pay in the 30^th instalment.

Given: first instalment =10000 and it increases the instalment by 500 every month.

So it form an AP with first term is 10000, common difference 500 and number of instalment is 30

Formula Used: T_n = a + (n - 1)d

(Where a is first term, T_n is nth term and d is common difference of given AP)

T_n = a + (n - 1)d T_n = 10000 + (30 - 1)500 T_n = 10000 + 29500

T_n = 10000 + 14500 T_n = 24,500

So, he will pay 24,500 in the 30^th instalment.

To Find: The sum of 25 terms of the given AP series.

Sum of n terms of an AP with first term a and common difference d is given by

Here, a = 17, n = 23 and d = - 5

= - 874

Sum of 23 terms of the AP IS - 874.

To find: Sum of 16 terms of the AP

Given:

First term = 6

Common difference =

S_n = 16

The sum of first 16 terms of the series is 16

To Find: The sum of 25 terms of the given AP series.

Sum of n terms of an AP with first term a and common difference d is given by

Here,

a = √2, n = 25, d = √2

= 25 × 13 × √2 = 325√2

Sum of 25 terms is 325√2.

To Find: The sum of 100 terms of the given AP series.

Sum of n terms of an AP with first term a and common difference d is given by

Here a = 0.6, n = 100, d = 0.01

= 50[1.2 + 0.99]

= 50 × 2.19

109.5
Sum of the series is 109.5

To Find: The sum of 20 terms of the given AP.

Sum of n terms of an AP with first term a and common difference d is given by

Here a = x + y, n = 20, d = - 2y

⇒S = 10[2x + 2y + 19( - 2y)] = 10[2x + 2y - 38y] = 10[2x - 36y]

⇒S = 20[x - 18y]

Sum of the series is 20(x - 18y).

To Find: The sum of n terms of the given AP.

Sum of n terms of an AP with first term a and common difference d is given by

Here a = x - y, d = 2x - y

The sum of the series is

To Find: The sum of the given series.

The nth term of an AP series is given by

t_n = a + (n - 1)d

⇒191 = 2 + (n - 1)3

⇒3(n - 1) = 189

⇒n - 1 = 63

⇒n = 64

Therefore,

= 32 × 193 = 6176

The sum of the series is 6176.

To Find: The sum of the given series.

Sum of the series is given by

Where n is the number of terms , a is the first term and l is the last term

Here a = 101, l = 43 ,n = 30

= 15 × 144 = 2160

The sum of the series is 2160.

Note: The sum of the series is already provided in the question. The solution to find x is given below.

Let there be n terms in the series.

x = 1 + (n - 1)3

= 3n - 2

Let S be the sum of the series

⇒n[1 + 3n - 2] = 1430

⇒n + 3n² - 2n = 1430

⇒3n² - n - 1430 = 0

Applying Sri Dhar Acharya formula, we get

⇒ n = 22 as n cannot be a fraction

Therefore x = 3 × 22 - 2 = 64

The value of x is 64

To Find: The value of x, i.e. the last term.

Given: The series and its sum.

The series can be written as x, (x + 3), …, 16, 19, 22, 25

Let there be n terms in the series

25 = x + (n - 1)3
3(n - 1) = 25 - x
x = 25 - 3(n - 1) = 28 - 3n

Let S be the sum of the series

⇒n[28 - 3n + 25] = 224

⇒n(53 - 3n) = 224

⇒3n² - 53n + 224 = 0

⇒n = 7 as n cannot be a fraction.

Therefore, x = 28 - 3n = 28 - 3(7) = 28 - 21 = 7

The value of x is 7.

Given: The sum of first n terms.

To Find: The r^th term.

Let the first term be a and common difference be d

Put n = 1 to get the first term
a = S₁ = 3 + 2 = 5
Put n = 2 to get a + (a + d)
2a + d = 12 + 4 = 16
10 + d = 16
d = 6
t_r = a + (r - 1)d
t �_r = 5 + (r - 1)6 = 5 + 6r - 6 = 6r - 1

The r^th term is given by 6r - 1.

To Find: The sum of n terms of an AP

Given: The r^th term.

The r^th term of the series is given by

Sum of the series is given by sum upto n terms of t_r

To Find: Last term of the AP.

Let the number of terms be n.

⇒n[54 - 3n + 3] = - 60

⇒3n² - 57n - 60 = 0

Either n = 20 or n = - 1 (n cannot be negative)

Therefore n = 20

Also,

, where l is the last term.

⇒ - 30 = 270 + 10l

⇒l = - 30

The last term is - 30.

To Find: Number of terms required

Let the number of terms be n.

⇒n[52 - 5n + 5] = 22
⇒n(57 - 5n) = 11 × 2 = 11[57 - 5(11)]
⇒n = 11

11 terms are required to give the sum 11.

To Find: Number of terms required to make the sum 78.

Here a = 18, d = - 2

Let n be the number of terms required to make the sum 78.

⇒78 × 2 = 36n - 2n² + 2n

⇒n² - 19n + 78 = 0

⇒n² - 6n - 13n + 78 = 0

⇒n(n - 6) - 13(n - 6) = 0

(n - 13)(n - 6) = 0

either n = 13 or n = 6

Explanation: Since the given AP is a decreasing progression where a_{n - 1}>a_n,it is bound to have negative values in the series. S_n is maximum for n = 9 or n = 10 since T₁₀ is 0(S₁₀ = S₉ = S_max = 90). The sum of 78 can be attained by either adding 6 terms or 13 terms so that negative terms from T₁₁ onward decrease the maximum sum to 78.

To Find: Number of terms required to make the sum of the AP 300.

Let the first term of the AP be a and the common difference be d

Here a = 20,

⇒300 × 6 = n[120 - 2(n - 1)]

⇒n[ - 2n + 122] = 6 × 300

⇒n( - n + 61) = 3 × 300

⇒n = 36 or 25

Explanation: Since the given AP is a decreasing progression where a_{n - 1}>a_n,it is bound to have negative values in the series. S_n is maximum for n = 30 or n = 31(S₃₀ = S₃₁ = S_max = 310). The sum of 300 can be attained by either adding 25 terms or 36 terms so that negative terms decrease the maximum sum to 300.

Wrong question. It will be 7n + 5 instead of 7n – 5.

Given: Ratio of sum of n terms of 2 AP’s

To Prove: 6^th terms of both AP’S are equal

Let us consider 2 AP series AP₁ and AP₂.

Putting n = 1, 2, 3… we get AP₁ as 12,19,26… and AP₂ as 22,27,32….

So, a_{1 =} 12, d_{1 =} 7 and a_{2 =} 22, d_{2 =} 5

For AP₁

S₆ = 12 + (6 - 1)7 = 47

For AP₂

S₆ = 22 + (6 - 1)5 = 47

Therefore their 6^th terms are equal.

Hence proved.

Given: Ratio of sum of n^th terms of 2 AP’s

To Find: Ratio of their 11^th terms

Let us consider 2 AP series AP₁ and AP₂.

Putting n = 1, 2, 3… we get AP₁ as 8, 15 22… and AP₂ as 31, 35, 39….

So, a_{1 =} 8, d_{1 =} 7 and a_{2 =} 31, d_{2 =} 4

For AP₁

S₆ = 8 + (11 - 1)7 = 87

For AP₂

S₆ = 31 + (11 - 1)4 = 81

Required ratio =

To Find: The sum of all odd integers from 1 to 201.

The odd integers form the following AP series:

1,3,5….201

First term = a = 1

Common difference = d = 2

Last term = 201

Let the number of terms be n

⇒1 + 2(n - 1) = 201

⇒n - 1 = 100

⇒n = 101

= 101 × 101 = 10201

The sum of all odd integers from 1 to 201 is 10201.

To Find: The sum of all even integers between 101 and 199.

The even integers form the following AP series -

102, 104, …, 198

It is and AP series with a = 102 and l = 198.

198 = 102 + (n - 1)2

⇒96 = (n - 1)2

⇒48 = n - 1

⇒n = 49

Now,

The sum of all even integers between 101 and 199 is 7350.

To Find: Sum of all integers between 101 and 500 divisible by 9

The integers between 101 and 500 divisible by 9 are 108, 117, 126,…, 495(Add 9 to 108 to get 117, 9 to 117 to get 126 and so on).

Let a be the first term and d be the common difference and n be the number of terms of the AP

Here a = 108, d = 9, l = 495

⇒a + (n - 1)d = 495

⇒108 + 9(n - 1) = 495

⇒12 + (n - 1) = 55

⇒n = 55 - 11 = 44

Now,

⇒S = 22[216 + 387] = 22[603] = 13266

Sum of all integers divisible by 9 between 100 and 500 is 13266.

The integers between 100 and 600 divisible by 5 and leaves remainder 2 are 102, 107, 112, 117,…, 597.

To Find: Sum of the above AP

Here a = 102, d = 5, l = 597

a + (n - 1)d = 597

⇒102 + 5(n - 1) = 597

⇒ (n - 1) = 99

⇒n = 100

Now,

⇒S = 50[204 + 495] = 50 × 699 = 34950

The sum of all such integers is 34950.

To Find: AP

Given: Sum of first 7 terms = 10
Sum of next 7 terms = 17

According to the problem,

Sum of first 14 terms of the given AP is 10 + 17 = 27.

So we can say

Solving the equations we get 14a + 42d = 20…(i) and

14a + 91d = 27… (

subtracting (i)from (ii)we get 49d = 7

⇒a = 1

To Find: m

Given: Sum of n terms, m^th term

Put n = 1 to get the first term

So a_{1 =} 3 + 5 = 8

Put n = 2 to get the sum of first and second term

So a₁ + a₂ = 12 + 10 = 22

So a_{2 =} 14

Common difference = 14 - 8 = 6

T_{n =} a + (n - 1)d = 8 + (n - 1)6 = 6n + 2

Now 6m + 2 = 164

Or m = 27

The value of m is 27.

To Find: The sum of all natural numbers from 1 to 100 which are divisible by 4 or 5.

A number divisible by both 4 and 5 should be divisible by 20
which is the LCM of 4 and 5.

Sum of numbers divisible by 4 OR 5 = Sum of numbers divisible by 4 + Sum of numbers divisible by 5 - Sum of numbers divisible by both 4 and 5.

Sum of numbers divisible by 4 = 4 + 8 + 12 + …100
= 4(1 + 2 + 3 + …25)
Sum of numbers divisible by 5 = 5 + 10 + 15 + 20 + …100
= 5(1 + 2 + 3 + ..20)

Sum of numbers divisible by 20 = 20 + 40 + 60...100
= 20(1 + 2 + 3 + 4 + 5) = 20 × 15 = 300
Required sum = 1800 + 1050 - 300 = 2550

Sum of numbers which are divisible by 4 or 5 is 2550

Let the first term be a and common difference be d

To Find: d

Given: Sum of n terms of AP =

⇒2a + (n - 1)d = 2P + (n - 1)Q
⇒2(a - P) = (n - 1)(Q - d)
Put n = 1 to get the first term as sum of 1 term of an AP is the term itself.

⇒P = a
⇒ (n - 1)(Q - d) = 0
For n not equal to 1 Q = d

Common difference is Q.

Let the first term of the AP be a and the common difference be d

Given: S_m = m²p and S_n = n²p

To prove: S_p = p³

According to the problem

⇒2a + (m - 1)d = 2mp
⇒2a + (n - 1)d = 2np
Subtracting the equations we get,

(m - n)d = 2p(m - n)

Now m is not equal to n

So d = 2p
Substituting in 1^st equation we get

2a + (m - 1)(2p) = 2mp
⇒a = mp - mp + p = p

Hence proved.

Let the carpenter take n days to finish the job.

To Find: n

He builds 5 frames on day 1, 7 on day 2, 9 on day 3 and so on.

So it forms an AP 5, 7, 9, 11,… and so on.

We need to find the number of terms in this AP such that the sum of the AP will be equal to 192

Given: Sum of AP = 192

⇒n(n + 8) = 192 × 2 = 16 × 24
⇒n = 16

He finishes the job in 16 days.

Given:

Interior angles of a polygon are in A.P

Smallest angle = a = 52°

Common difference = d = 8°

Let the number of sides of a polygon = n

Angles are in the following order

52°, 52° + d, 52° + 2d, ........, 52° + (n - 1) ×d

Sum of n terms in A.P = s

Sum of angles of the given polygon is

Hint:

Sum of interior angles of a polygon of n sides is

Therefore,

180n - 360 = 52n + n (n - 1) ×4

4n² + 48n = 180n - 360

4n² - 132n + 360 = 0

n² - 33n + 90 = 0

(n - 3)(n - 30) = 0

n = 3 &n = 30

∴ It can be a triangle or a 30 sided polygon.

The number of sides of the polygon is 3 or 30.

A circle is divided into n sectors.

Given,

Angles are in A.P

Smallest angle = a = 8°

Largest angle = l = 72°

Final term of last term of an A.P series is l = a + (n - 1)×d

So,

72° = 8° + (n - 1) ×d

(n - 1) ×d = 64°

Sum of all angles of all divided sectors is

Sum of n terms of A.P whose first term and the last term are known is

Where nis the number of terms in A.P.

So,

n(40°) = 360°

n

n = 9

From equations (1) & (2) we get,

(9 - 1) ×d = 64°

8×d = 64°

d

d = 8°

The circle is divided into nine sectors whose angles are in A.P with a common difference of 8°.

Angle in fifth sector is a + (5 - 1) ×d = 40°

∴n = 9

The angle in the fifth sector = 40°.

Hint:

Distances between trees and well are in A.P.

Given:

The distance of well from its nearest tree is 10 metres

Distance between each tree is 5 metres.

So,

In A.P

The first term is 10 metres and the common difference is 5 metres.

a = 10 & d = 5

The distances are in the following order

10, 15, 20… (30 terms)

The farthest tree is at a distance of a + (30 - 1) ×d

l = 10 + (29) ×5

L = 155metres.

Total distance travelled by the Gardner = 2×Sum of all the distances of 30 trees from the well.

Sum of distances of all the 30 trees is

Sum

= 15×165 metres

= 2475 metres.

Total distance travelled by the Gardner is 2 × 2475metres.

∴The total distance travelled by the Gardner is 4950 metres.

Given :

Two cars start together from the same place and move in the same direction.

The first car moves with a uniform speed of 60km/hr.

The second car moves with 48km/hr in the first hour and increases the speed by 1 km each succeeding hour.

Let the cars meet at n hours.

Distance travelled the first car in n hours = 60×n

Distance travelled by the second car in n hours is

Tip: -

When the cars meet the distances travelled by cars are equal.

96 + (n - 1) = 120

n = 25

∴ The two cars meet after 25 hoursfrom their start and overtake the first car.

Given:

The amount that is to be paid to buy a scooter = 44000

The amount that he paid by cash = ₹8000

Remaining balance = ₹36000

Annual instalment = ₹4000 + [email protected]% on the unpaid amount

Thus, our instalments are 7600, 7200, 6800…….

Total number of instalments

= 9

So our instalments are 7600, 7200, 6800 ... up to 9 terms.

Hint: - All our instalments are in A.P with a common difference of 400.

Here

First term, a = 7200

Common difference = d = 7200 - 7600

d = - 400

Number of terms = 9

Sum of all instalments

= 54000

Hence,

The total cost of the scooter = amount that is paid earlier + amount paid in 9 instalments.

= 8000 + 54000

= 62000

∴The total cost paid by Arun = 62000

Given: -

An initial salary that will be given = ₹26000

There will be an automatic increase of ₹250 per month from the very next month and thereafter.

Hint: - In the given information the salaries he receives are in A.P.

Let the number of the month is n.

Initial salary = a = ₹26000

Increase in salary = common difference = d = ₹250

i. Salary for the 10^th month,

n = 10,

Salary = a + (n - 1)×d

= 26000 + (10 - 1)×250

= 28250

∴ Salary for the 10^th month = ₹28250

ii. Total earnings during the first year = sum off all salaries received per month.

Total earnings =

Here n = 12.

Total earnings

= 6×(42000 + 2750)

= 268500

Total earnings during the first year = ₹268500

Given: -

Amount saved by a man in 20 years is Rs.660000.

Let the amount saved by him in the first year be.

In every succeeding year, he saves Rs.2000 more than what he saved in the previous year.

Increment of saving of the year when compared last year is Rs.2000

Hint: - The above information looks like the savings are in Arithmetic Progression.

Amount saved in first year = a

Common difference = d = ₹2000

Total number of years = n = 20

The total amount saved in 20 years is ₹660000

Sum of n terms in an A.P

a = 14000

∴ In the first year, he saved ₹14000.

Given: -

Initially let the work can be completed in ndays when 150 workers work on every day.

However every day 4 workers are being dropped from the work so that work took 8 more days to be finished.

Finally, it takes (n + 8) days to finish the works.

Work equivalent when 150 workers work without being dropped = 150×n

Work equivalent when workers are dropped day by day = 150 + (150 - 4) + (150 - 8) + ...... + (150 - 4(n + 8)).

So,

150×n = 150 + (150 - 4) + ........ + (150 - 4×(n + 8))

150×n = 150×n + 150×8 - 4×(1 + 2 + 3 + ...... + (n + 8))

(n + 8)(n + 9) = 600

n² + 17n - 528 = 0

n = - 33 or n = 16

Since the number of days cannot be negative, n = 16.

∴ In 24 days the work is completed.

A Man saves some amount of money every year.

In the first year, he saves Rs.4000.

In the next year, he saves Rs.5000.

Like this, he increases his savings by Rs.1000 ever year.

Given a total amount of Rs. 85000 is saved in some ‘n’ years.

According to the above information the savings in every year are in Arithmetic Progression.

First year savings = a = Rs.4000

Increase in every year savings = d = Rs.1000

Total savings (s_n) = Rs.85000

Sum of n terms in A.P

n² + 7×n - 170 = 0

(n + 17) ×(n - 10) = 0

n = - 17 or n = 10

Since the number of years cannot be negative, n = 10.

After 10 years his savings will become Rs.85000.

Given: -

Total debt = Rs.36000

A man pays this debt in 40 annual instalments that forms an A.P.

After annual instalments, that man dies leaving one - third of the debt unpaid.

So,

Within 30 instalments he pays two - thirds of his debt.

Sum of n terms in an Arithmetic Progression =

He has to pay 36000 in 40 annual instalments,

Where,

a = amount paid in the first instalment,

d = difference between two Consecutive instalments.

He paid two – a third of the debt in 30 instalments,

From equations (1) & (2)we get,

a = 510 & d = 20

∴The value of the first instalment is Rs.510.

Hint: - In the question it is mentioned that the production increases by a fixed number every year.

So it is an A.P. (a₁, a₂, a₃, a₄, ........a_{n - 1}, a_n).

Given: -

The 3^rd year production is 6000 units

So,

a₃ = 6000

We know that a_n = a + (n - 1) ×d

a₃ = a + (3 - 1)×d

6000 = a + 2d

The 7^th year production is 7000 units

So,

a₇ = 7000

a₇ = a + (7 - 1)×d

7000 = a + 6d

From equations (1)&(2) we get,

6000 - 2d = 7000 - 6d

4×d = 1000

d = 250

From equations (1)&(2) we get,

a = 5500

i. Production in the first year = a = 5500

∴5500 units were produced by the manufacturer of TV sets in the first year.

ii. Production in the 10^th year = a₁₀ = a + (10 - 1)×d

a₁₀ = 5500 + (9) ×250

= 7750

∴7750 units were produced by the manufacturer of TV sets in the 10^th year.

iii. Total production in seven years = a₁ + a₂ + a₃ + a₄ + a₅ + a₆ + a₇

s₇ = 43750

∴A total of 16, 250 units was produced by the manufacturer in 7 years.

-

Given: -

The amount that is to be paid to buy a tractor = ₹180000.

An amount that he paid by cash = ₹90000.

Remaining balance = ₹90000

Annual instalment = ₹9000 + interest @12% on unpaid amount.

Thus, our instalments are 19800, 18720, 17640…….

Total number of instalments =

=

= 10

So our instalments are 19800, 18720, 17640 ... upto 10 terms.

All our instalments are in A.P with a common difference d.

Here

First term(a) = 19800

Common difference = d = 18720 - 19800

d = - 1080

Number of terms is 10

Sum of all instalments

= 149400

Hence,

The total cost of the scooter = amount that is paid earlier + amount paid in 10 instalments.

= 90000 + 149400

∴The total cost paid by the farmer = ₹239400

(i) 9 and 19

To find: Arithmetic mean between 9 and 19

The formula used: Arithmetic mean between

We have 9 and 19

(ii) 15 and -7

To find: Arithmetic mean between 15 and -7

The formula used: Arithmetic mean between

We have 15 and -7

(iii) -16 and -8

To find: Arithmetic mean between -16 and -8

The formula used: Arithmetic mean between

We have -16 and -8

To find: Four arithmetic means between 4 and 29

Formula used: (i) , where, d is the common difference

n is the number of arithmetic means

(ii) A_n = a + nd

We have 4 and 29

Using Formula,

Using Formula, A_n = a + nd

First arithmetic mean, A₁ = a + d

= 4 + 5

= 9

Second arithmetic mean, A₂ = a + 2d

= 4 + 2(5)

= 4 + 10

= 14

Third arithmetic mean, A₃ = a + 3d

= 4 + 3(5)

= 4 + 15

= 19

Fourth arithmetic mean, A₄ = a + 4d

= 4 + 4(5)

= 4 + 20

= 24

Ans) The four arithmetic means between 4 and 29 are 9, 14, 19 and 24

To find: Three arithmetic means between 23 and 7

Formula used: (i) , where, d is the common difference

n is the number of arithmetic means

(ii) A_n = a + nd

We have 23 and 7

Using Formula,

Using Formula, A_n = a + nd

First arithmetic mean, A₁ = a + d

= 23 + (-4)

= 19

Second arithmetic mean, A₂ = a + 2d

= 23 + 2(-4)

= 23 + (-8)

= 15

Third arithmetic mean, A₃ = a + 3d

= 23 + 3(-4)

= 23 + (-12)

= 11

Ans) The three arithmetic means between 23 and 7 are 19, 15 and 11

To find: Six arithmetic means between 11 and -10

Formula used: (i) , where, d is the common difference

n is the number of arithmetic means

(ii) A_n = a + nd

We have 11 and -10

Using Formula,

Using Formula, A_n = a + nd

First arithmetic mean, A₁ = a + d

= 11 + (-3)

= 8

Second arithmetic mean, A₂ = a + 2d

= 11 + 2(-3)

= 11 + (-6)

= 5

Third arithmetic mean, A₃ = a + 3d

= 11 + 3(-3)

= 11 + (-9)

= 2

Fourth arithmetic mean, A₄ = a + 4d

= 11 + 4(-3)

= 11 + (-12)

= -1

Fifth arithmetic mean, A₅ = a + 5d

= 11 + 5(-3)

= 11 + (-15)

= -4

Sixth arithmetic mean, A₆ = a + 6d

= 11 + 6(-3)

= 11 + (-18)

= -7

Ans) The six arithmetic means between 11 and -10 are 8, 5, 2, - 1, -4 and -7.

To find: The value of n

Given: (i) The numbers are 9 and 27

(ii) The ratio of the last mean to the first mean is 2 : 1

Formula used: (i) , where, d is the common difference

n is the number of arithmetic means

(ii) A_n = a + nd

We have 9 and 27,

Using Formula,

Using Formula, A_n = a + nd

First mean i.e.,

… (i)

Last mean i.e.,

A_n … (ii)

The ratio of the last mean to the first mean is 2 : 1

Substituting the value of A₁ and A_n from eqn. (i) and (ii)

⇒ 27n + 9 = 18n + 54

⇒ 9n = 45

⇒ n = 5

Ans) The value of n is 5

To find: The number of arithmetic means

Given: (i) The numbers are 16 and 65

(ii) 5^th arithmetic mean is 51

Formula used: (i) , where, d is the common difference

n is the number of arithmetic means

(ii) A_n = a + nd

We have 16 and 65,

Using Formula,

Using Formula, A_n = a + nd

Fifth arithmetic mean, A₅ = a + 5d

A₅ = 51 (Given)

Therefore,

⇒ 245 = 35n + 35

⇒ 210 = 35n

⇒ n = 6

The number of arithmetic means are 6.

Using Formula,

d = 7

Using Formula, A_n = a + nd

First arithmetic mean, A₁ = a + d

= 16 + 7

= 23

Second arithmetic mean, A₂ = a + 2d

= 16 + 2(7)

= 16 + 14

= 30

Third arithmetic mean, A₃ = a + 3d

= 16 + 3(7)

= 16 + 21

= 37

Fourth arithmetic mean, A₄ = a + 4d

= 16 + 4(7)

= 16 + 28

= 44

Fifth arithmetic mean, A₅ = a + 5d

= 16 + 5(7)

= 16 + 35

= 51

Sixth arithmetic mean, A₆ = a + 6d

= 16 + 6(7)

= 16 + 42

= 58

Ans) The six arithmetic means between 1 and 65 are 23, 30, 37, 44, 51 and 58.

To find: Five numbers between 11 and 29, which are in A.P.

Given: (i) The numbers are 11 and 29

Formula used: (i) A_n = a + (n-1)d

Let the five numbers be A₁, A_2, A_3, A₄ and A₅

According to question 11, A₁, A_2, A_3, A_4, A₅ and 29 are in A.P.

We can see that the number of terms in this series is 7

For the above series:-

a = 11 , n=7

A₇ = 29

Using formula, A_n = a + (n-1)d

⇒ A₇ =11 + (7-1)d = 29

⇒ 6d = 29 – 11

⇒ 6d = 18

⇒ d = 3

We can see from the definition that A₁, A_2, A_3, A₄ and A₅ are five arithmetic mean between 11 and 29, where d = 3, a = 11

Therefore, Using formula of arithmetic mean i.e. A_n = a + nd

A₁ = a + nd

= 11 + 3

= 14

A₂ = a + nd

= 11 + (2)3

= 17

A₃ = a + nd

= 11 + (3)3

= 20

A₄ = a + nd

= 11 + (4)3

= 23

A₅ = a + nd

= 11 + (5)3

= 26

Ans) 14, 17, 20, 23 and 26 are the required numbers.

To prove: ratio of sum of m arithmetic means between the two numbers to the sum of n arithmetic means between them is m:n

Formula used: (i) , where, d is the common difference

n is the number of arithmetic means

(ii) , Where n = Number of terms

a = First term

l = Last term

Let the first series of arithmetic mean having m arithmetic means be,

a, A₁, A_2, A₃ … A_m, l

In the above series we have (m + 2) terms

⇒ l = a + (m + 2 – 1)d

⇒ l = a + (m + 1)d … (i)

In the above series A₁ is second term

⇒ A₁ = a + (2-1)d

= a + d

In the above series A_m is the (m+1)^th term

⇒ A_m = a + (m+1-1)d

= a + md

Now, A₁ + A_m = a + d + a + md

= a + a + (m+1)d

= a + l [From eqn (i)]

Therefore, A₁ + A_m = a + l … (ii)

For the sum of arithmetic means in the above series:-

First term = A_1, Last term = A_m, No. of terms = m

Using Formula,

From eqn. (ii)

Let the second series of arithmetic mean having n arithmetic means be,

a, A₁, A_2, A₃ … A_n, l

In the above series we have (n + 2) terms

⇒ l = a + (n + 2 – 1)d

⇒ l = a + (n + 1)d … (iii)

In the above series A₁ is second term

⇒ A₁ = a + (2-1)d

= a + d

In the above series A_n is the (n+1)^th term

⇒ A_n = a + (n+1-1)d

= a + nd

Now, A₁ + A_n = a + d + a + nd

= a + a + (n+1)d

= a + l [ From eqn (iii) ]

Therefore, A₁ + A_n = a + l … (iv)

For the sum of arithmetic means in the above series:-

First term = A_1, Last term = A_n, No. of terms = n

Using Formula,

From eqn. (iv)

There,

Hence Proved

(i) (a – c)² = 4(a – b)(b – c)

To prove: (a – c)² = 4(a – b)(b – c)

Given: a, b, c are in A.P.

Proof: Since a, b, c are in A.P.

⇒ c – b = b – a = common difference

⇒ b – c = a – b … (i)

And, 2b = a + c (a, b, c are in A.P.)

⇒ 2b – c = a … (ii)

Taking LHS = (a – c)²

= ( 2b – c – c )² [from eqn. (ii)]

= ( 2b – 2c )²

= 4( b – c )²

= 4( b – c ) ( b – c )

= 4( a – b ) ( b – c ) [b–c = a–b from eqn. (i)]

= RHS

Hence Proved

(ii) a² + c² + 4ac = 2(ab + bc + ca)

To prove: a² + c² + 4ac = 2(ab + bc + ca)

Given: a, b, c are in A.P.

Proof: Since a, b, c are in A.P.

⇒ 2b = a + c

… (i)

Taking RHS = 2(ab + bc + ca)

Substituting value of b from eqn. (i)

= a² + c² + 4ac

= LHS

Hence Proved

(iii) a³ + c³ + 6abc = 8b³

To prove: a³ + c³ + 6abc = 8b³

Given: a, b, c are in A.P.

Formula used: (a+b)³ = a³ + 3ab(a+b) + b³

Proof: Since a, b, c are in A.P.

⇒ 2b = a + c … (i)

Cubing both side,

⇒ 8b³ = a³ + 3ac(a+c) + c³

⇒ 8b³ = a³ + 3ac(2b) + c³ [a+c = 2b from eqn. (i)]

⇒ 8b³ = a³ + 6abc + c³

On rearranging,

a³ + c³ + 6abc = 8b³

Hence Proved

To prove: (a + 2b – c)(2b + c – a)(c + a – b) = 4abc.

Given: a, b, c are in A.P.

Proof: Since a, b, c are in A.P.

⇒ 2b = a + c … (i)

Taking LHS = (a + 2b – c) (2b + c – a) (c + a – b)

Substituting the value of 2b from eqn. (i)

= (a + a + c – c) (a + c + c – a) (c + a – b)

= (2a) (2c) (c + a – b)

Substituting the value of (a + c) from eqn. (i)

= (2a) (2c) (2b – b)

= (2a) (2c) (b)

= 4abc

= RHS

Hence Proved

(i) (b + c – a), (c + a – b), (a + b – c) are in AP.

To prove: (b + c – a), (c + a – b), (a + b – c) are in AP.

Given: a, b, c are in A.P.

Proof: Let d be the common difference for the A.P. a,b,c

Since a, b, c are in A.P.

⇒ b – a = c – b = common differnce

⇒ a – b = b – c = d

⇒ 2(a – b) = 2(b – c) = 2d … (i)

Considering series (b + c – a), (c + a – b), (a + b – c)

For numbers to be in A.P. there must be a common difference between them

Taking (b + c – a) and (c + a – b)

Common Difference = (c + a – b) - (b + c – a)

= c + a – b – b – c + a

= 2a – 2b

= 2(a – b)

= 2d [from eqn. (i)]

Taking (c + a – b) and (a + b – c)

Common Difference = (a + b – c) - (c + a – b)

= a + b – c – c – a + b

= 2b – 2c

= 2(b – c)

= 2d [from eqn. (i)]

Here we can see that we have obtained a common difference between numbers i.e. 2d

Hence, (b + c – a), (c + a – b), (a + b – c) are in AP.

(ii) (bc – a²), (ca – b²), (ab – c²) are in AP.

To prove: (bc – a²), (ca – b²), (ab – c²) are in AP.

Given: a, b, c are in A.P.

Proof: Let d be the common difference for the A.P. a,b,c

Since a, b, c are in A.P.

⇒ b – a = c – b = common differnce

⇒ a – b = b – c = d … (i)

Considering series (bc – a²), (ca – b²), (ab – c²)

For numbers to be in A.P. there must be a common difference between them

Taking (bc – a²) and (ca – b²)

Common Difference = (ca – b²) – (bc – a²)

= [ca – b² – bc + a²]

= [ca – bc + a² – b²]

= [c (a – b) + (a + b) (a – b)]

= [(a – b ) (a + b + c)]

a – b = d, from eqn. (i)

⇒ [(d) (a + b + c)]

Taking (ca – b²) and (ab – c²)

Common Difference = (ab – c²) – (ca – b²)

= [ab – c² – ca + b²]

= [ab – ca + b² – c²]

= [a (b – c) + (b – c) (b + c)]

= [(b – c) (a + b + c)]

b – c = d, from eqn. (i)

⇒ [(d) (a + b + c)]

Here we can see that we have obtained a common difference between numbers i.e. [(d) (a + b + c)]

Hence, (bc – a²), (ca – b²), (ab – c²) are in AP.

(i) are in A.P.

To prove: are in A.P.

Given: are in A.P.

Proof: are in A.P.

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Multiplying the A.P. with ( a + b + c )

are also in A.P.

If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

Substracting the above A.P. with 1

, are also in A.P.

, are also in A.P.

, are also in A.P.

Hence Proved

(ii) are in A.P.

To prove: are in A.P.

Given: are in A.P.

Proof: are in A.P.

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Multiplying the A.P. with ( a + b + c )

are also in A.P.

If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

Substracting the above A.P. with 2

, are also in A.P.

, are also in A.P.

, are also in A.P.

Hence Proved

To prove: a²(b + c), b²(c + a), c²(a + b) are in A.P.

Given: are in A.P.

Proof:are in A.P.

are in A.P.

are in A.P.

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Multiplying the A.P. with (abc)

, are in A.P.

are in A.P.

⇒ [(a²c + a²b)], [ab² + b²c], [c²b + ac²] are in A.P.

On rearranging,

⇒ [a²(b + c)], [b²(c + a)] , [c²(a + b)] are in A.P.

Hence Proved

To prove: are in A.P.

Given: a, b, c are in A.P.

Proof: a, b, care in A.P.

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Multiplying the A.P. with (ab + bc + ac)

, are in A.P.

Multiplying the A.P. with

, are in A.P.

, are in A.P.

If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

Substracting the A.P. with 1

, are in A.P.

, are in A.P.

On rearranging

, are in A.P.

Hence Proved

Given: S_n = (2n² + 3n)

To find: find common difference

Put n = 1 we get

S_{1 =} 5 OR we can write

a = 5 …equation 1

Similarly put n = 2 we get

S₂ = 14 OR we can write

2a + d = 14

Using equation 1 we get

d = 4

Given : 9 × (9^th term) = 13 × (13^th term)

To prove: 22^nd term is 0

9 × (a + 8d) = 13 × (a + 12d)

9a + 72d = 13a + 156d

- 4a = 84d

a = - 21d …..Equation 1

Also 22^nd term is given by

a + 21d

Using equation 1 we get

- 21d + 21d = 0

Hence proved 22^nd term is 0.

Given: S_n = qn² , S_m = qm²

To prove: S_q = q³

Put n = 1 we get

a = q …… equation 1

Put n = 2

2a + d = 4q ……equation 2

Using equation 1 and 2 we get

d = 2q

So

S_q = q³

Hence proved.

let the three AM be x₁,x₂,x₃.

So new AP will be

6,x₁,x₂,x_3, - 6

Also - 6 = 6 + 4d

d = - 3

x₁ = 3

x₂ = 0

x₃ = - 3

Given :9^th term is 0

To prove: 29^th term is double the 19^th term

a + 8d = 0

a = - 8d

29^th term is

a + 28d

⟹ 20d

19^th term is

a + 18d

⟹ 10d

Hence proved 29^th term is double the 19^th term

To find: number of terms in AP

Also

d = 16 – 13

d = 3

Also

43 = 13 + n × 3 – 3

So

n = 11

To find: 8^th term from the end

d = 9 - 7

d = 2

Also

201 = 7 + n × 2 – 2

n = 98

So 8^th term from end will be

7 + 90 × 2

⟹ 187

the first 2 digit number divisible by 7 is 14, and the last 2 digit number divisible by 7 is 98, so it forms AP with common difference 7

14,…,98

98 = 14 + (n - 1) × 7

n = 22

Given: 7^th term is 34 and 8^th term is 64

To find: find its 18^th term

34 = a + 6d ………….equation1

64 = a + 12d ………… equation2

Subtract equation1 from equation2 we get

d = 5

Put in equation1 we get

a = 4

So 18^th term is

4 + 17 × 5 = 89

To find: 10^th common term between the APs

Common difference of 1^st series = 4

Common difference of 2^nd series = 5

LCM of common difference will give us a common difference of new series

⟹ 5 × 4

⟹ 20

The first term of new AP will be 11, so the 10^th = term of this series is

⟹ 11 + 20 × 9

⟹ 191

Given: the sum of its terms is 36, the first and last terms of an AP are 1 and 11.

To find: the number of terms

Sum of AP using first and last terms is given by

36 × 2 = n (1 + 11)

n = 6

Given: p^th term is q and (p + q)^th term is 0.

To prove: q^th term is p.

p^th term is given by

q = a + (p - 1) × d……equation1

(p + q)^th term is given by

0 = a + (p + q - 1) × d

0 = a + (p - 1) × d + q × d

Using equation1

0 = q + q × d

d = - 1

Put in equation1 we get

a = q + p - 1

q^th term is

⟹ q + p - 1 + (q - 1) × ( - 1)

⟹p

Hence proved.

To find: the value of n.

We can write it as

3n² + 7 × n - 370 = 0

Therefore n = 37/3, 10

Rejecting 37/3 we get n = 10

even natural numbers are

2, 4, 6, 8…..

n odd natural numbers are given by

3,5,7,9,…….

Given: the sum of n terms of an AP is.

To find: common difference.

put n = 1 we get

First term =

Put n = 2 we get

First term + second term = 2 × a + 2 × b

Second term =

Therefore common difference will be

Second term – first term

Common difference = 2a

Given: sums of n terms of two APs are in ratio (2n + 3) : (3n + 2)

To find: find the ratio of their 10^th terms.

For the sum of n terms of two APs is given by

Or we can write it as

For 10^th term put

n = 19

Therefore the ratio of the 10^th term will be