Question

Question 1
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

Let the first term, common differences and the number of terms of an AP are a, d and n, respectively.
Sum of first n terms of an AP, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]\cdots \left(i\right)$
$\therefore$ Sum of first five terms of an AP, $S_5=\frac{5}{2}[2a+(5-1\left)d\right]$ [from Eq.(i)]
$=\frac{5}{2}(2a+4d)=5(a+2d)$
$\Rightarrow S_5=5a+10d$
And, sum of first seven terms of an AP, $S_7=\frac{7}{2}[2a+(7-1\left)d\right]$
$=\frac{7}{2}[2a+6d]=7(a+3d)$
$\Rightarrow S_7=7a+21d$
Now, by given condition,
$S_5+S_7=167$
$\Rightarrow 5a+10d+7a+21d=167$
$\Rightarrow 12a+31d=167$ ...(ii)
Given that, sum of first ten terms of this AP is 235.
$\therefore S_{10}=235$
$\Rightarrow \frac{10}{2}[2a+(10-1\left)d\right]=235$
$\Rightarrow 5(2a+9d)=235$
$\Rightarrow 2a+9d=47$ ...(iii)
Solving (ii) and (iii) gives,
$12a+54d=28212a+31d=167---\underset{–}{}23d=115d=5$
Now, put the value of d in Eq. (iii), we get,
$2a+9\left(5\right)=47\Rightarrow 2a+45=47$
$\Rightarrow 2a=47-45=2\Rightarrow a=1$
Sum of first twenty terms of this AP, $S_{20}=\frac{20}{2}[2a+(20-1\left)d\right]$
$=10[2\times (1)+19\times (5\left)\right]=10(2+95)$
$=10\times 97=970$