Question

For what value of $p$ are $2p+1,13,5p-3,$ three consecutive terms of an A.P. ?

• A. 2
• B. $-2$
• C. 4
• D. $-4$

Answer 1

The correct option is C 4

Given:

$2p+1,13,5p-3$ are in A.P.

$\therefore 2\times 13=2p+1+5p-3$ $[2b=c+a]$

$\Rightarrow 26=7p-2$

$\Rightarrow 7p=28$

$\therefore p=4$