Question

The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16, find the sum of its first 10 terms. [CBSE 2015]

Answer 1

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

$$\begin{gathered} a_{13}=4\times a_3\left(\mathrm{Given}\right) \\ \Rightarrow a+12d=4\left(a+2d\right)\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+12d=4a+8d \\ \Rightarrow 3a=4d.....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} a_5=16\left(\mathrm{Given}\right) \\ \Rightarrow a+4d=16.....\left(2\right) \end{gathered}$$

Solving (1) and (2), we get

$$\begin{gathered} a+3a=16 \\ \Rightarrow 4a=16 \\ \Rightarrow a=4 \end{gathered}$$

Putting a = 4 in (1), we get

$$\begin{gathered} 4d=3\times 4=12 \\ \Rightarrow d=3 \end{gathered}$$

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

$$\begin{gathered} S_{10}=\frac{10}{2}\left[2\times 4+\left(10-1\right)\times 3\right] \\ =5\times \left(8+27\right) \\ =5\times 35 \\ =175 \end{gathered}$$

Hence, the required sum is 175.