For what value of $p$, are $( 2p-1),\ 7$ and $\frac{11}{2}p$ three consecutive terms of an A.P.?

Given: $( 2p-1),\ 7$ and $\frac{11}{2}p$ are three consecutive terms of an A.P.

To do: To find the value of $p$.

Solution: 

As given terms, $( 2p-1),\ 7$ and $\frac{11}{2}p$, to be in A.P. 

$7-( 2p-1)=\frac{11}{2}p-7$

$\Rightarrow 7-2p+1=\frac{11p-14}{2}$

$\Rightarrow 8-2p=\frac{11p-14}{2}$

$\Rightarrow 2( 8-2p)=11p-14$

$\Rightarrow 16-4p=11p-14$

$\Rightarrow -11p-4p=-16-14$

$\Rightarrow -15p=--30$

$\Rightarrow p=\frac{-30}{-15}$

$\Rightarrow p=2$

Thus, for $p=2$, $( 2p-1),\ 7$ and $\frac{11}{2}p$ are three consecutive terms of an A.P.