In an AP:
Given $a = 7, a_{13} = 35$, find $d$ and $S_{13}$.

Given:

In an A.P., $a = 7, a_{13} = 35$

To do:

We have to find $d$ and $S_{13}$.

Solution:

We know that,

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

$a_{13}=35$

$a+12 d=35$

$7+12 d=35$

$12d=35-7$

$d=\frac{28}{12}$

$d=\frac{7}{3}$

$S_{13}=\frac{13}{2}[a+a_{13}]$

$=\frac{13}{2}[7+35]$

$=\frac{13}{2}(42)$

$=13 \times 21$

$=273$