Question

If a,b,c are in A.P. prove that a 2 +c 2 +4ac=2(ab+bc+ca)

Answer 1

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Answer 2

Proved that, a² + c² + 4ac = 2(ab + bc + ca).

Step-by-step explanation:

Given that a, b, and c are in A.P.

Hence, if the common difference is d, then, b = a + d and c = a + 2d.

So. we have to prove that a² + c² + 4ac = 2(ab + bc + ca).

Now, LHS = a² + c² + 4ac

= a² + (a + 2d)² + 4a(a + 2d)

= a² + a² + 4ad + 4d² + 4a² + 8ad

= 6a² + 12ad + 4d²

Now, RHS = 2(ab + bc + ca)

= 2[a(a + d) + (a + d)(a + 2d) + a(a + 2d)]

= 2[a² + ad + a² + 3ad + 2d² + a² + 2ad]

= 6a² + 12ad + 4d²

Hence, proved that, a² + c² + 4ac = 2(ab + bc + ca).