Question

The ${11}^{th}$ term of an AP exceeds its $4^{th}$ term by $14$. Find the common difference.

Answer 1

$4th$ term $=a+3d$

$11th$ term $=a+10d$

According to the given equation,

$a+3d+14=a+10d$

$7d=14$

$d=2$

Hence, this is the answer.