For what value of $p$ are $2 p+1,13,5 p-3$ are three consecutive terms of an A.P.?
Here, we are given three terms,
First term (a1) =
Second term (a2) =
Third term (a3) =
We need to find the value of p for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
$d=12-2 p$ $\ldots(1)$
Now, on equating (1) and (2), we get,
$12-2 p=5 p-16$
$5 p+2 p=16+12$
Therefore, for $p=4$, these three terms will form an A.P.