Question

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167 . If the sum of the first ten terms of this AP is 235 . Find the sum of its first twenty terms ? ​

Answer 1

Sum of n terms of an AP:

Sn  =  n/2(2a + (n - 1)d)

Sum of first 5 terms (n = 5) :-

⇒ S  =  5/2(2a + (5 - 1)d)

⇒ S = 5/2(2a + 4d)

⇒ S = (10a + 20d)/2 ....... (1)

Sum of first 7 terms (n = 7) :-

⇒ S  =  7/2(2a + (7 - 1)d)

⇒ S = 7/2(2a + 6d)

⇒ S = (14a + 42d)/2 ....... (2)

Given: (1) + (2) = 167

⇒ (10a + 20d)/2 + (14a + 42d)/2 = 167

⇒ (10a + 20d + 14a + 42d)/2 = 167

⇒ 24a + 62d = 167 x 2

⇒ 24a + 62d = 334 ÷ 2

⇒ 12a + 31d = 167 ..........(3)

Also,

Given:- Sum of first 10 terms of this AP is 235.

Sum of first 10 terms (n = 10) :-

⇒  10/2(2a + (10 - 1)d)

⇒ 5(2a + 9d)

⇒ (10a + 45d) = 235 ÷ 5

⇒ 2a + 9d = 47 .......... (4)

Multiplying equation (4) with 6 :-

⇒ 2a + 9d = 47 x 6

⇒ 12a + 54d = 282 .......... (5)

Now, (eq.5) - (eq.3)

12a + 54d = 282

12a + 31d = 167

-------------------------

23d = 115

d = 115/23

d = 5

Substituting d = 1 in (eq. 5)

12a + 54d = 282

12a + 54(5) = 282

12a + 270 = 282

12a = 12

a = 1

Sum of first 20 terms (n = 20) :-

⇒ S  =  20/2(2(1) + (20 - 1)5)

⇒ S = 10(2 + 19(5))

⇒ S = 10 x 97

⇒ S = 970

Answer 2

Answer:

• The sum of the first 20 terms = 970

Explanation:

Let, the first term of AP be 'a' and

common difference of AP be 'd' then

Considering the formula for the sum 'Sₙ' of first 'n' terms of an AP with the first term 'a' and common difference 'd'

Sₙ = (n/2) (2a + (n-1)d)

Given that,

→ Sum of first five terms of an AP + Sum of first seven terms of an AP = 167

→ S₅ + S₇ = 167

→ (5/2) (2a + (5-1)d) + (7/2) (2a + (7-1)d) = 167

→ (5/2) (2a + 4d) + (7/2) (2a + 6d) = 167

→ 5 (a + 2d) + 7 (a + 3d) = 167

→ 5a + 10d + 7a + 21d = 167

→ 12a + 31d = 167    ....equation(1)

Now, Also given that

→ Sum of the first 10 terms = 235

→ S₁₀ = 235

→ (`0/2) (2a + (10-1)d) = 235

→ 5 (2a + 9d) = 235

→ 10a + 45d = 235

[ dividing by 5 both sides ]

→ 2a + 9d = 47

[ multiplying by 6 both sides ]

→ 12 a + 54d = 282   ....equation(2)

Subtracting equation(1) from equation(2)

→ 12a + 54d - (12a + 31d) = 282 - 167

→ 12a - 12a + 54d - 31d = 282 - 167

→ 23d = 115

→ d = 5

Putting the value of 'd' in equation(1)

→ 12a + 31d = 167

→ 12a + 31(5) = 167

→ 12a + 155 = 167

→ 12a = 167 - 155

→ 12a = 12

→ a = 1

Now, we need to find the sum of first 20 terms of that AP (n = 20) , so

→ S₂₀ = (20/2) (2a + (n-1)d)

→ S₂₀ = 10 (2(1) + (20-1)(5))

→ S₂₀ = 10 (2 + 95)

→ S₂₀ = 970

Therefore,

• The sum of the first 20 terms of AP is 970.