Show that (a − b)2, (a2 + b2) and (a + b)2 are in A.P.

Question:

Show that $(a-b)^{2},\left(a^{2}+b^{2}\right)$ and $(a+b)^{2}$ are in A.P.

Solution:

Here, we are given three terms and we need to show that they are in A.P.,

First term $\left(a_{1}\right)=(a-b)^{2}$

Second term $\left(a_{2}\right)=\left(a^{2}+b^{2}\right)$

 

Third term $\left(a_{3}\right)=(a+b)^{2}$

So, in an A.P. the difference of two adjacent terms is always constant. So, to prove that these terms are in A.P. we find the common difference, we get,

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

$d=\left(a^{2}+b^{2}\right)-(a-b)^{2}$

$d=a^{2}+b^{2}-\left(a^{2}+b^{2}-2 a b\right)$

$d=a^{2}+b^{2}-a^{2}-b^{2}+2 a b$

$d=2 a b$.............(1)

Also,

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

$d=(a+b)^{2}-\left(a^{2}+b^{2}\right)$

$d=a^{2}+b^{2}+2 a b-a^{2}-b^{2}$

$d=2 a b \ldots \ldots$ (2)

Now, since in equations (1) and (2) the values of are equal, we can say that these terms are in A.P. with 2ab as the common difference.

Hence proved

 

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