Question

Find the sum of 32 terms of an A.P. whose third terms is 1 and 6 th term is $-11$

Answer 1

Let the first term and common difference of an $A.P$ be $a,d$ respectively

Given Third term is $a+2d=1\cdot \left(1\right)$

$6$ th term is $a+5d=-11\cdots \left(2\right)$

$\left(2\right)-\left(1\right)⟹3d=-12⟹d=-4$

Substitute $d=-4$ in $\left(1\right)$

$⟹a-8=1⟹a=8+1=9$

Sum of $32$ terms is $\frac{32}{2}\left(2\right(9)+(32-1\left)\right(-4\left)\right)=16(18-124)=-16\times 106=-1696$