Let there be an A.P. with first term ‘$a$’, common difference '$d$'. If $a_n$ denotes its $n^{th}$ term and $S_n$ the sum of first $n$ terms, find.$n$ and $a_n$, if $a = 2, d = 8$ and $S_n = 90$.

Given:

In an A.P., first term $=a$ and common difference $=d$.

$a_n$ denotes its $n^{th}$ term and $S_n$ the sum of first $n$ terms.

To do:

We have to find $n$ and $a_n$, if $a = 2, d = 8$ and $S_n = 90$.

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,

$a_n=2+(n-1)8$

$=2+8n-8$

$=8n-6$........(i)

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

$90=\frac{n}{2}[4+8n-8]$        (From (i))

$90(2)=n(8n-4)$

$180=4n(2n-1)$

$n(2n-1)=45$

$2n^2-n-45=0$

$2n^2-10n+9n-45=0$

$2n(n-5)+9(n-5)=0$

$(2n+9)(n-5)=0$

$n=5$ or $2n=-9$ which is not possible as $n$ cannot be negative

$\therefore n=5$

This implies,

$a_n=8(5)-6$

$=40-6$

$=34$

Therefore, $n=5$ and $a_n=34$.