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Question
For what value of p are 2p + 1, 13, 5p − 3 are three consecutive terms of an A.P.?
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Solution
Here, we are given three terms,
First term (a1) = 2p + 1
Second term (a2) = 13
Third term (a3) = 5p - 3
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 = a2 - a1
d = 13 - (2p + 1)
d = 13 - 2p - 1
d = 12 - 2p .............(1)
Also,
d = a3 - a2
d = (5 p - 3) - 13
d = 5p - 3 - 13
d = 5p - 16 ..................(2)
Now, on equating (1) and (2), we get,
12 - 2p = 5p - 16
5p + 2p = 16 + 12
7p = 28
`p = 28/7`
p = 4
Therefore, for p = 4 , these three terms will form an A.P.
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