Q. 7 B5.0( 3 Votes )

If a, b, c are in A.P., prove that

(b + c)2 — a2, (c + a)2 — b2, (a + b)2 — c2 are in A.P.

Answer :

Given: a, b, c are in AP


Since, a, b, c are in AP, we have a + c = 2b …(i)


Now, (b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c2 will be in A.P


If (b + c – a)(b + c + a), (c + a – b)(c + a + b), (a + b – c)(a + b + c) are in AP


i.e. if b + c – a, c + a – b, a + b – c are in AP


[dividing by (a + b + c)]


if (b + c – a) + (a + b – c) = 2(c + a – b)


if 2b = 2(c + a – b)


if b = c + a – b


if a + c = 2b which is true by (i)


Hence, (b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c2 are in A.P


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