Question

The sum of first 10 terms of an AP is −150 and the sum of its next 10 terms is −550. Find the AP. [CBSE 2010]

Answer 1

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

$$\begin{gathered} S_{10}=-150\left(\mathrm{Given}\right) \\ \Rightarrow \frac{10}{2}\left(2a+9d\right)=-150\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ \Rightarrow 5\left(2a+9d\right)=-150 \\ \Rightarrow 2a+9d=-30.....\left(1\right) \end{gathered}$$

It is given that the sum of its next 10 terms is −550.

Now,

S₂₀ = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms = −150 + (−550) = −700

$$\begin{gathered} \therefore S_{20}=-700 \\ \Rightarrow \frac{20}{2}\left(2a+19d\right)=-700 \\ \Rightarrow 10\left(2a+19d\right)=-700 \\ \Rightarrow 2a+19d=-70.....\left(2\right) \end{gathered}$$

Subtracting (1) from (2), we get

$$\begin{gathered} \left(2a+19d\right)-\left(2a+9d\right)=-70-\left(-30\right) \\ \Rightarrow 10d=-40 \\ \Rightarrow d=-4 \end{gathered}$$

Putting d = −4 in (1), we get

$$\begin{gathered} 2a+9\times \left(-4\right)=-30 \\ \Rightarrow 2a=-30+36=6 \\ \Rightarrow a=3 \end{gathered}$$

Hence, the required AP is 3, −1, −5, −9, ... .