In an AP:
Given $a = 8, a_n = 62, S_n = 210$, find $n$ and $d$.

Given:

In an A.P., $a = 8, a_n = 62, S_n = 210$

To do:

We have to find $n$ and $d$.

Solution:

We know that,

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

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

This implies,

$a_n=8+(n-1)d$

$62=8+(n-1)d$

$62-8=(n-1)d$

$54=(n-1)d$

$(n-1)d=54$........(i)

$S_n=\frac{n}{2}[2 \times 8+(n-1)d]$

$210=\frac{n}{2}[16+54]$        (From (i))

$210(2)=n(70)$

$3(2)=n$

$n=6$

$\therefore (6-1)d=54$

$5d=54$

$d=\frac{54}{5}$

Therefore, $n=6$ and $d=\frac{54}{5}$.