Question

In an AP, it is given that S₅ + S₇ = 167 and S₁₀ = 235, then find the AP, where S_n denotes the sum of its first n terms. [CBSE 2015]

Answer 1

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

$$\begin{gathered} S_5+S_7=167 \\ \Rightarrow \frac{5}{2}\left(2a+4d\right)+\frac{7}{2}\left(2a+6d\right)=167\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ \Rightarrow 5a+10d+7a+21d=167 \\ \Rightarrow 12a+31d=167.....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} S_{10}=235 \\ \Rightarrow \frac{10}{2}\left(2a+9d\right)=235 \\ \Rightarrow 5\left(2a+9d\right)=235 \\ \Rightarrow 2a+9d=47 \end{gathered}$$

Multiplying both sides by 6, we get

$12a+54d=282.....\left(2\right)$

Subtracting (1) from (2), we get

$$\begin{gathered} 12a+54d-12a-31d=282-167 \\ \Rightarrow 23d=115 \\ \Rightarrow d=5 \end{gathered}$$

Putting d = 5 in (1), we get

$$\begin{gathered} 12a+31\times 5=167 \\ \Rightarrow 12a+155=167 \\ \Rightarrow 12a=167-155=12 \\ \Rightarrow a=1 \end{gathered}$$

Hence, the AP is 1, 6, 11, 16, ... .