Question

There are $60$ terms in an A.P. of which the first term is $8$ and the last term is $185.$ The ${31}^{st}$ term is

• A. $56$
• B. $94$
• C. $85$
• D. $98$

Answer 1

The correct option is D $98$

Let $d$ be the common difference, then,

${60}^{th}$ term $=8+59d=185$
$\Rightarrow 59d=177$

$\Rightarrow d=3$

$\Rightarrow {31}^{st}$ term $=8+30\times 3=98$