Question:
If $(2 p-1), 7$ and $3 p$ are in $\mathrm{AP}$, find the value of $p$.
Solution:
Let $(2 p-1), 7$ and $3 p$ be three consecutive terms of an AP.
Then $7-(2 p-1)=3 p-7$
$\Rightarrow 5 p=15$
$\Rightarrow p=3$
$\therefore$ When $p=3,(2 p-1), 7$ and $3 p$ form three consecutive terms of an AP.