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प्रश्न
If a, b, c and d are in proportion, prove that: `abcd [(1/a^2 + 1/b^2 + 1/c^2 + 1/d^2]` = a2 + b2 + c2 + d2
उत्तर
∵ a, b, c, d are in proportion
`a/b = c/d` = k(say)
a = bk, c = dk.
L.H.S. = `abcd (1/a^2 + 1/b^2 + 1/c^2 + 1/d^2)`
= `bk.b.dk.d [1/(b^2k^2) + 1/b^2 + 1/(d^2k^2) + 1/d^2]`
= `k^2b^2d^2 [(d^2 + d^2k^2 + b^2 + b^2k^2)/(b^2d^2k^2)]`
= d2(1 + k2) + b2(1 + k2)
= (1 + k2)(b2 + d2)
R.H.S. = a2 + b2 + c2 + d2
= b2k2 + b2 + d2k2 + d2
= b2(k2 + 1) + d2(k2 + 1)
= (k2 + 1)(b2 + d2)
∴ L.H.S. = R.H.S.
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