Question:
Find the value of p for which the numbers 2p – 1, 3p + 1, 11 are in AP. Hence, find the numbers.
Solution:
$2 p-1,3 p+1,11$ are in $\mathrm{AP}$
A.P has common difference
$d=a_{2}-a_{1}$
Here, $a_{1}=2 p-1, a_{2}=3 p+1$ and $a_{3}=11$
$a_{2}-a_{1}=a_{3}-a_{2}$
$3 p+1-(2 p-1)=11-3 p-1$
$p+2=10-3 p$
$\Rightarrow p=2$
And the numbers are:
$2(2)-1,3(2)+1,11$
$\Rightarrow 3,7,11$ are the numbers