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.

#### Solution

Let the first term, common difference and the number of terms of an AP are a, d and n, respectively.

∵ Sum of first n terms of an AP

`S_n = n/2 [2a + (n - 1_d]` .....(i)

∴ Sum of first five terms of an AP,

`S_5 = 5/2[2a + (5 - 1)d]`.........[From equation (i)]`

⇒ `S_5 = 5a + 10d` .....(ii)

And sum of first seven terms of an AP,

`S_7 = 7/2[2a + (7 - 1)d]`

= `7/2[2a + 6d]`

= `7(a + 3d)`

⇒ `S_7 = 7a + 21d` .....(iii)

Now, by given condition,

S_{5} + S_{7} = 167

⇒ 5a + 10 d + 7a + 21 d = 16 7

⇒ 12a + 31d = 167 ........(iv)

Given that, sum of first ten terms of this AP is 235.

∴ S_{10} = 235

⇒ `10/2 [2a + (10 - 1)d]` = 235

⇒ 5(2a + 9d) = 235

⇒ 2a + 9d = 47 …(v)

On multiplying equation (v) by 6 and then subtracting it into equation (iv), we get

23d = 115

⇒ d = 5

Now, put the value of d in equation (v), we get

2a + 9(5) = 47

⇒ 2a + 45 = 47

⇒ 2a = 47 – 45 = 2

⇒ a = 1

Sum of first twenty terms of this AP,

`S_20 = 20/2 [2a + (20 - 1)d]`

= 10[2 × (1) + 19 × (5)]

= 10(2 + 95)

= 10 × 97

= 970

Hence, the required sum of its first twenty terms is 970.