Question

Three numbers form an increasing G.P if the middle number is doubled, then the new numbers are in A.P. The common ratio of the G.P is:​

Answer 1

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$let \: the \: three \: numbers \: in \: g.p \: be \: a \: \frac{a}{r} \: ar \\ \\ then \: \frac{a}{r} \: 2a \: ar \: are \: in \: a.p \: as \: given \\ \\ we \: know \: that \: 2b = ac \\ \\ therefore \: 2(2a) = a(r + \frac{1}{r} ) \\ 4a = a( \frac{ {r}^{2} + 1}{r} ) \\ 4r = {r}^{2} + 1 \\ {r}^{2} - 4r + 1 \\ r = 2 + \sqrt{3} \: and \: r = 2 - \sqrt{3}$

Answer 2

SOLUTION

Let a, ar, ar² are in GP (r>1).

According to this question, a, 2ar, ar² in A.P.

=) 4ar= a+ ar²

=)r² -4r+1 = 0

$= > r = \frac{4 ± \sqrt{16 - 4} }{2} \\ \\ = > r = \frac{4 ± \sqrt{12} }{2} \\ \\ = > r = \frac{4 ± 2 \sqrt{3} }{2} \\ \\ = > r = 2 ± \sqrt{3} \: \: \: \: \: \: (taking \: 2 \: common)$

=)r= 2+√3 or r=2-√3

=)r= 2+√3 [So, A.P is increasing]

Hope it helps ☺️