Question

The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms. [CBSE 2015]

Answer 1

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

$$\begin{gathered} a_{16}=5\times a_3\left(\mathrm{Given}\right) \\ \Rightarrow a+15d=5\left(a+2d\right)\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+15d=5a+10d \\ \Rightarrow 4a=5d.....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} a_{10}=41\left(\mathrm{Given}\right) \\ \Rightarrow a+9d=41.....\left(2\right) \end{gathered}$$

Solving (1) and (2), we get

$$\begin{gathered} a+9\times \frac{4a}{5}=41 \\ \Rightarrow \frac{5a+36a}{5}=41 \\ \Rightarrow \frac{41a}{5}=41 \\ \Rightarrow a=5 \end{gathered}$$

Putting a = 5 in (1), we get

$$\begin{gathered} 5d=4\times 5=20 \\ \Rightarrow d=4 \end{gathered}$$

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

$$\begin{gathered} S_{15}=\frac{15}{2}\left[2\times 5+\left(15-1\right)\times 4\right] \\ =\frac{15}{2}\times \left(10+56\right) \\ =\frac{15}{2}\times 66 \\ =495 \end{gathered}$$

Hence, the required sum is 495.