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प्रश्न
If a, b, c, d are in continued proportion, prove that:
`((a -b)/c + (a - c)/b)^2 - ((d - b)/c + (d - c)/b)^2 = (a - d)^2 (1/c^2 - 1/b^2)`
उत्तर
a, b, c, d are in continued proportion
∴ `a/b = b/c = c/d` = k(say)
∴ c = dk, b = ck = dk, k = dk2
a = bk = dk2. k = dk3
L.H.S. = `((a -b)/c + (a - c)/b)^2 - ((d - b)/c + (d - c)/b)^2`
= `((dk^3 - dk^2)/(dk) + (dk^3 - dk)/(dk^2))^2 - ((d - dk^2)/(dk) + (d - dk)/(dk^2))^2`
= `((dk^2(k - 1))/(dk) + (dk(k^2 - 1))/(dk^2))^2 - ((d(1 - k^2))/(dk) + (d( 1 - k^2))/(dk^2))^2`
= `((k(k - 1) + (k^2 - 1))/k)^2 - ((1 - k^2)/k + (1 - k)/k^2)^2`
= `((k^2(k - 1) + (k^2 - 1))/k)^2 - ((k (1- k^2) + 1 - k)/k^2)^2`
= `((k^3 - 1)^2)/k^2 - (-k^3 + 1)^2/k^4`
= `(k^3 - 1)^2/k^2 - (1 - k^3)^2/k^4`
= `((k^3 - 1)/k^2)^2 ((1 - 1)/k^2)`
= `((k^3 - 1)^2(k^2 - 1))/k^4`
= `((k^3 - 1)^2(k^2 - 1))/k^4`
R.H.S. = `(a – d)^2(1 / c^2 - 1/b^2)`
= `(dk^3 - d)^2(1 / (d^2k^2) - (1)/(d^2k^4))`
= `d^2(k^3 - 1)^2((k^2 - 1)/(d^2k^4))`
= `((k^3 - 1)^2(k^2 - 1))/k^4`
∴ L.H.S. = R.H.S.
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