In an AP:
Given $a_3 = 15, S_{10} = 125$, find $d$ and $a_{10}$.

Given:

In an A.P., $a_3 = 15, S_{10} = 125$

To do:

We have to find $d$ and $a_{10}$.

Solution:

We know that,

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

$a_{3}=15$

$a+2d=15$

$a=15-2d$......(i)

$S_{10}=\frac{10}{2}[2a+(10-1)d]$

$=5[2(15-2d)+9d]$

$=5[30-4d+9d]$

$=5(30+5 d)$

$125=5(30+5d)$

$\frac{125}{5}=30+5d$

$25-30=5d$

$-5=5 d$

$d=-1$

This implies,

$a=15-2(-1)$

$=15+2$

$=17$

$a_{10}=a+(10-1)(-1)$

$=17+9(-1)$

$=17-9$

$=8$