You may test if the given series is an arithmetical progression using the following relation between three consecutive terms of arithmetical progression, `a_k, a_(k+1), a_(k+2),` such that:
`a_(k+1) = (a_k + a_(k+2))/2`
Considering `a_k = (a+b)^2, a_(k+1) = a^2 + b^2` and `a_(k+2) = (a-b)^2` and reasoning by analogy, you need to test the next relation, such that:
`a^2 + b^2 = ((a+b)^2 + (a-b)^2)/2`
Expanding the squares, yields:
`a^2 + b^2 = (a^2 + 2ab + b^2 + a^2 - 2ab + b^2)/2`
Reducing duplicate terms, yields:
`a^2 + b^2 = (2a^2 + 2b^2)/2`
Factoring out 2 yields:
`a^2 + b^2 = 2(a^2 + b^2)/2`
Reducing duplicate factors, yields:
`a^2 + b^2 = a^2 + b^2 `
Hence, using the equation that relates three consecutive terms of an arithmetical progression yields that `(a+b)^2, a^2 + b^2, (a-b)^2` are the terms of arithmetical progression.
To show that they are AP we have to prove that thet have common difference from term to term. Or Just see that Secon term -first term = 3rd term -2nd term orT2-T1=T3-T2.
So, T2-T1 gives a^2+b^2-(a+b)^2 = -2ab
T3-T2 gives:(a-b)^2-(a^2+b^2) = -2ab.
So, the given terms in the order keep acommon difference -2ab and that proves they are in AP.
The given series has three terms
(a+b)^2,(a)^2 + (b)^2, (a-b)^2,
We can expand the first term, that is (a+b)^2, as:
(a)^2 + 2ab + (b)^2.
Similarly we can expand the third term, that is (a-b)^2, as:
(a)^2 - 2ab + (b)^2.
Substituting these values of the first and third terms in the series it becomes:
(a)^2 + 2ab + (b)^2, (a)^2 + (b)^2, (a)^2 - 2ab + (b)^2,
We can see that in this series the second term is obtained by subtracting 2ab from the first term. and the third term is obtained by subtracting the same quantity from second term. Thus this series is a series in which the difference between any two consecutive terms id 2ab. Therefore this series is an arithmetical progression.