Question

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koi is ki coding de do na ​

Answer 1

Since a, b, c, d are in HP, 1/a, 1/b, 1/c and 1/d are in AP.

D = 1/b−1/a = 1/c−1/b = 1/d−1/c

where D is the common difference of the AP: 1/a, 1/b, 1/c.

Now,

ab = a−b/D

bc = b−c/D

cd = c−d/D

Adding all we get,

ab+bc+cd = (a−b+b−c+c−d)/D = (a-d)/D --------(1)

Now,

D = 1/3(1/d-1/a)

D = 1/3(a-d)/ad

(a-d)/D = 3ad ------------(2)

putting (2) in (1) we get,

ab+bc+cd = 3ad (Ans)