Question:
If $(2 p+1), 13,(5 p-3)$ are in AP, find the value of $p$.
Solution:
Let $(2 p+1), 13,(5 p-3)$ be three consecutive terms of an AP.
Then $13-(2 p+1)=(5 p-3)-13$
$\Rightarrow 7 p=28$
$\Rightarrow p=4$
$\therefore$ When $p=4,(2 p+1), 13$ and $(5 p-3)$ form three consecutive terms of an AP.