Question

If the $8^{th}$ term of an A.P. is $31$ and the ${15}^{th}$ term is $16$ more than the ${11}^{th}$ term, find the A.P.

Answer 1

Let $a$ be the first term and $d$ be the common difference of the A.P.

We have,
$a_8=31$ and $a_{15}=16+a_{11}$

$⟹a+7d=31$ and $14d=16+a+10d$

$⟹a+7d=31$ and $4d=16$

$⟹a+7d=31$ and $d=4$

$⟹a+7\times 4=31⟹a+28=31⟹a=3$
Hence, the A.P. is $a,a+d,a+2d,a+3d,....$ i.e. $3,7,11,15,19,....$