If a- b- c are in continued proportion then show that a-b 2b-c 2-a 2-b 2 b 2-c 2

Given: #160;ab=bcLet: #160;ab=bc=k #8756; #160;a=bk #160;and #160;b=ckNow,LHS=a+b2b+c2=bk+b2ck+c2=b2k+12c2k+12=b2c2RHS=a2+b2b2+c2=b2k2+b2c2k2+c2=b2c ...

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Standard X

Mathematics

Proportion

If a, b, c ar...

Question

If a, b, c are in continued proportion then show that $\frac{(a + b)^{2}}{(b + c)^{2}} = \frac{a^{2} + b^{2}}{b^{2} + c^{2}} .$

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Solution

$Given : \frac{a}{b} = \frac{b}{c} Let : \frac{a}{b} = \frac{b}{c} = k ∴ a = b k and b = c k Now, LHS = \frac{{(a + b)}^{2}}{{(b + c)}^{2}} = \frac{{(b k + b)}^{2}}{{(c k + c)}^{2}} = \frac{b^{2} {(k + 1)}^{2}}{c^{2} {(k + 1)}^{2}} = \frac{b^{2}}{c^{2}} RHS = \frac{a^{2} + b^{2}}{b^{2} + c^{2}} = \frac{b^{2} k^{2} + b^{2}}{c^{2} k^{2} + c^{2}} = \frac{b^{2}}{c^{2}} LHS = RHS ∴ \frac{{(a + b)}^{2}}{{(b + c)}^{2}} = \frac{a^{2} + b^{2}}{b^{2} + c^{2}}$

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If a, b, c are in continued proportion then show that (a + b)2(b +c)2 = a2 + b2b2 + c2.

Given: ab=bcLet: ab=bc=k∴ a=bk and b=ckNow,LHS=a+b2b+c2=bk+b2ck+c2=b2k+12c2k+12=b2c2RHS=a2+b2b2+c2=b2k2+b2c2k2+c2=b2c2LHS=RHS∴a+b2b+c2=a2+b2b2+c2

If a, b, c are in continued proportion then show that $\frac{(a + b)^{2}}{(b + c)^{2}} = \frac{a^{2} + b^{2}}{b^{2} + c^{2}} .$