In an AP:
Given $a_{12} = 37, d = 3$, find $a$ and $S_{12}$.

Given:

In an A.P., $a_{12} = 37, d = 3$

To do:

We have to find $a$ and $S_{12}$.

Solution:

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$a_{12}=37$

$a+11 d=37$

$a+11(3)=37$

$a=37-33$

$a=4$

Therefore,

$S_{12}=\frac{12}{2}[a+a_{12}]$

$=6[4+37]$

$=6 \times 41$

$=246$