Question

10. If a, b and c are in continued proportion, prove that
3a2-4ab + 5b2 /3b2 -4bc +5c2=a/c​

Answer 1

EXPLANATION.

If a, b, c are in continued proportion.

Prove that :

$\sf \displaystyle \frac{3a^{2} - 4ab + 5b^{2} }{3b^{2} - 4bc + 5c^{2} } = \frac{a}{c}$

As we know that,

If a, b, c are in continued proportion then,

⇒ a/b = b/c.

$\sf \displaystyle \frac{a}{b} = \frac{b}{c} = k \ (say)$

$\sf \displaystyle b = ck. - - - - - (1)$

$\sf \displaystyle a = bk$

Put the value of b = ck in the expression, we get.

$\sf \displaystyle a = (ck)(k)$

$\sf \displaystyle a = ck^{2}. - - - - - (2)$

Put the value of a and b in the equation, we get.

$\sf \displaystyle \frac{3(ck^{2}) ^{2} - 4(ck^{2} )(ck) + 5(ck)^{2} }{3(ck)^{2} - 4(ck)c + 5c^{2} }$

$\sf \displaystyle \frac{3c^{2} k^{4} - 4c^{2}k^{3} + 5c^{2}k^{2} }{3c^{2} k^{2} - 4c^{2} k + 5c^{2} }$

$\sf \displaystyle \frac{c^{2}k^{2}(3k^{2} - 4k + 5) }{c^{2}(3k^{2} - 4k + 5) }$

$\sf \displaystyle \frac{c^{2}k^{2} }{c^{2} } = k^{2} = \frac{a}{c}$

Hence proved.