Find the value of p for which the numbers 2p – 1, 3p + 1, 11 are in AP.
Question:

Find the value of for which the numbers 2p – 1, 3+ 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

 

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