Three numbers are in an increasing geometric progression with common ratio r . If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d . If the

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Three numbers are in an increasing geometric progression with common ratio r . If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d . If the fourth term of GP is 3 r 2 , then r 2 - d is equal to :

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Three numbers are in an increasing geometric progression with common ratio $r .$

Let first term is $a$

$\Rightarrow \frac{a}{r}, a, a r \in G . P$

Given that if middle term is doubled

$\Rightarrow \frac{a}{r}, 2 a, a r \in A . P$

If $a, b, c \in A . P \Rightarrow 2 b = a + c$
$4 a = a r + \frac{a}{r}$

$4 = r + \frac{1}{r}$

$r^{2} + 1 = 4 r$

$r^{2} - 4 r + 1 = 0$

If $a x^{2} + b x + c = 0 \Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a c}$

$\Rightarrow r = \frac{4 \pm \sqrt{12}}{2} = 2 + \sqrt{3}, 2 - \sqrt{3}$

But it is an increasing $G . P$

$∴ r = 2 + \sqrt{3}$

Given that Fourth term of $G . P \Rightarrow t_{4} = a r^{2} = 3 r^{2}$

$\Rightarrow a = 3 .$

Common difference of an $A . P = d = t_{2} - t_{1}$

$d = 2 a - \frac{a}{r} = a (2 - \frac{1}{r}) = 3 (2 - \frac{1}{2 + \sqrt{3}})$

$= 3 (2 - 2 + \sqrt{3}) = 3 \sqrt{3}$
Hence $r^{2} - d = (2 + \sqrt{3})^{2} - (3 \sqrt{3})$

$= 4 + 3 + 4 \sqrt{3} - 3 \sqrt{3}$

$= 7 + \sqrt{3}$

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