Question

The ${16}^{th}$ term of an AP is five times its third term. If its ${10}^{th}$ term is 41, then find the sum of its first fifteen terms.

Answer 1

Let the first term of AP be a and common differnce be d.
Given that ${16}^{th}$ term of an AP is five times its third term.

i.e., $a+(16-1)d=5[a+(3-1\left)d\right]$

$\Rightarrow a+15d=5[a+2d]$

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

$\Rightarrow 4a-5d=0....\left(i\right)$

Also given that, $a_{10}=41$

$\Rightarrow a+(10-1d)=41$

$\Rightarrow a+9d=41....\left(ii\right)$

On multiplying equation (ii) by 4, we get

4a+36d=164 ....(iii)

Subtracting equation (iii) from (i), we get

$\begin{array}{c}4a-5d=0\\ 4a+36d=164\\ \underset{–}{---}\\ -41d=-164\\ d=4\end{array}$

On putting the value of d in eq.(i), we get

$4a-5\times 4=0$

4a=20

a=5

Now, $S_{15}=\frac{15}{2}[2a+(15-1\left)d\right]$

$S_{15}=\frac{15}{2}(2\times 5+14\times 4)$

$=\frac{15}{2}2(5+14\times 4)$

=15(5+28)

=$15\times 33$

$S_{15}=495$

Hence,sum of first fifteen terms is 495.