Question

The sum of the first and the fifth term of an arithmetic progression is $26$ and the product of the second by the fourth term is $160$. Find the sum of the first six terms of the progression.

Answer 1

Series is in $AP$

$1^{st}$ and $5^{th}$ term sum$=26$

Let $a$ is the first term and $d$ is the common difference.

$a_r=a+(r-1)da+a+(5-1)d=262a+4d=26a+2d=13-\left(i\right)$

$2^{nd}$ and $4^{th}$ term product $=160$

$a=13-2d(a+d)(a+3d)=160-\left(ii\right)$

Putting value of $a$ from $\left(i\right)$ in $\left(ii\right)$

$(13-2d+d)(13-2d+3d)=160(13-d)(13+d)=160169-d^2=160d^2=9d=3$

Putting value of $d$ in $\left(i\right)$

$a=13-2\left(3\right)a=13-6a=7$

Sum of first $6$ terms

$=\frac{n}{2}(2a+(n-1\left)d\right)=\frac{6}{2}\left(2\right(7)+(6-1\left)\right(3\left)\right)=3(14+5(3\left)\right)=3(14+15)=3\left(29\right)=87$

Sum of first $6$ terms$=87$.