Question

If the $8$ term of an A.P is $31$ and the $15$th term is $16$ more than the $11$th term, then $23$rd term of the A.P is

• A. $88$
• B. $89$
• C. $95$
• D. $91$

Answer 1

Answer: (D)

$91$

$a_8=a+7d=31$
$a_{11}=a+10d$
$a_{15}=a+14d$
According to the question 11 th term is 16 more than 15 th term
then $a+14d=a+10d+16$
$4d=16$
$d=4$
$a+7d=31$
$a=31-7\times 4$
$a=31-28=3$
Then 23 term is $a_{23}=a+22d$
$\Rightarrow 3+22\times 4$
$\Rightarrow 3+88=91$