In an AP:
Given $d = 5, S_9 = 75$, find $a$ and $a_9$.

Given:

To do:

We have to find $a$ and $a_9$.

Solution:

We know that,

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

$S_{9}=\frac{9}{2}[2 a+(9-1) d]$

$75=\frac{9}{2}[2 a+8 d]$

$75 \times \frac{2}{9}=2a+8(5)$

$\frac{50}{3}=2 a+40$

$2a=\frac{50}{3}-40$

$2a=\frac{50-120}{3}$

$2a=\frac{-70}{3}$

$a=\frac{-35}{3}$

$a_{9}=a+8 d$

$=\frac{-35}{3}+8(5)$

$=\frac{-35}{3}+40$

$=\frac{-35+120}{3}$

$=\frac{85}{3}$