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.

Solution:

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

$a_{13}=4 \times a_{3}$                      (Given)

$\Rightarrow a+12 d=4(a+2 d) \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+12 d=4 a+8 d$

$\Rightarrow 3 a=4 d \quad \ldots(1)$

Also,

$a_{5}=16$                 (Given)

$\Rightarrow a+4 d=16 \quad \ldots$ (2)

Solving (1) and (2), we get

$a+3 a=16$

$\Rightarrow 4 a=16$

$\Rightarrow a=4$

Putting a = 4 in (1), we get

$4 d=3 \times 4=12$

$\Rightarrow d=3$

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

$S_{10}=\frac{10}{2}[2 \times 4+(10-1) \times 3]$

$=5 \times(8+27)$

$=5 \times 35$

 

$=175$

Hence, the required sum is 175 .