Question

The sum of the 4th and the 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44. Find the sum of its first 10 terms. [CBSE 2013C]

Answer 1

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

$$\begin{gathered} \therefore a_4+a_8=24\left(\mathrm{Given}\right) \\ \Rightarrow \left(a+3d\right)+\left(a+7d\right)=24\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 2a+10d=24 \\ \Rightarrow a+5d=12.....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} \therefore a_6+a_{10}=44\left(\mathrm{Given}\right) \\ \Rightarrow \left(a+5d\right)+\left(a+9d\right)=44\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 2a+14d=44 \\ \Rightarrow a+7d=22.....\left(2\right) \end{gathered}$$

Subtracting (1) from (2), we get

$$\begin{gathered} \left(a+7d\right)-\left(a+5d\right)=22-12 \\ \Rightarrow 2d=10 \\ \Rightarrow d=5 \end{gathered}$$

Putting d = 5 in (1), we get

$$\begin{gathered} a+5\times 5=12 \\ \Rightarrow a=12-25=-13 \end{gathered}$$

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

$$\begin{gathered} S_{10}=\frac{10}{2}\left[2\times \left(-13\right)+\left(10-1\right)\times 5\right] \\ =5\times \left(-26+45\right) \\ =5\times 19 \\ =95 \end{gathered}$$

Hence, the required sum is 95.