Question

Find the number of terms of the AP: $-12,-9,-6,...12$. If $1$ is added to each term of this AP, then find the sum of all terms of the AP thus obtained.

Answer 1

Given A.P is $-12,-9,-6,\cdots ,12$

If $1$ is added to each, we A.P becomes

$(-12+1),(-9+1),(-6+1)\cdots ,(12+1)$

Or $-11,-8,-5,\cdots ,13$

First Term, $a=-11$ and common difference is $3$

Last term, $l=a_n=13$

$a_n=a+(n-1)d$

$13=-11+(n-1)3$

$\Rightarrow 13=-11+3n-3$

$\Rightarrow 3n=13+11+3$

$\Rightarrow 3n=27$

$\Rightarrow n=9$

So $13$ is $9^{th}$ term

$S_n=\frac{n}{2}(a+l)$

$S_9=\frac{9}{2}(-11+13)=\frac{9}{2}\times 2=9$