Question

The first term, common difference and the last term of an $AP$ are $12,6$ and $252$ respectively. Find the sum of all the terms of this $AP$.

Answer 1

Given,

$a=12,d=6,a_n=252$

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

$252=12+(n-1)6$

$40=n-1$

$\therefore n=41$

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

$=\frac{41(12+252)}{2}$

$=5412$