For what value of p are 2p + 1, 13, 5p − 3

Question:

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

Solution:

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,

$d=a_{2}-a_{1}$

$d=13-(2 p+1)$

$d=13-2 p-1$

$d=12-2 p$ $\ldots(1)$

Also,

$d=a_{3}-a_{2}$

$d=(5 p-3)-13$

$d=5 p-3-13$

 

$d=5 p-16$...........(2)

Now, on equating (1) and (2), we get,

$12-2 p=5 p-16$

 

$5 p+2 p=16+12$

$7 p=28$

$p=\frac{28}{7}$

 

$p=4$

Therefore, for $p=4$, these three terms will form an A.P.

 

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