The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41,

Solution:

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

$a_{16}=5 \times a_{3}$                                      (Given)

$\Rightarrow a+15 d=5(a+2 d) \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+15 d=5 a+10 d$

$\Rightarrow 4 a=5 d \quad \ldots(1)$

Also,

$a_{10}=41$                               (Given)

$\Rightarrow a+9 d=41 \quad \ldots(2)$

Solving (1) and (2), we get

$a+9 \times \frac{4 a}{5}=41$

$\Rightarrow \frac{5 a+36 a}{5}=41$

$\Rightarrow \frac{41 a}{5}=41$

$\Rightarrow a=5$

Putting a = 5 in (1), we get

$5 d=4 \times 5=20$

$\Rightarrow d=4$

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

$S_{15}=\frac{15}{2}[2 \times 5+(15-1) \times 4]$

$=\frac{15}{2} \times(10+56)$

$=\frac{15}{2} \times 66$

$=495$

Hence, the required sum is 495.