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What are the two middle terms of four positive terms of GP whose sum is 40 and product is 729

Option: 1

1,3


Option: 2

3,9


Option: 3

9,27


Option: 4

1,9


Answers (1)

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As we learnt

Selection of terms in G.P.

If we have to take four terms of a GP, we take them as

\frac{a}{r^{3}},\frac{a}{r},ar,ar^{3}

 

Now,

Let the terms be

\frac{a}{r^{3}},\frac{a}{r},ar,ar^{3}

Thus product = a^{4}= 729\Rightarrow a=3^{\frac{6}{4}}= 3^{\frac{3}{2}}= 3\sqrt{3}

a\left (\frac{1}{r^{3}}+\frac{1}{r}+r+r^{3} \right )= 40;\:\\ r^{3} + \frac{1}{r^{3}} + r + \frac{1}{r} = \frac{40}{3\sqrt3} \\(r + \frac{1}{r})^3 - 3 (r + \frac{1}{r}) + (r + \frac{1}{r}) - \frac{40}{3\sqrt3} = 0 \\ Putting\: (r + \frac{1}{r}) = t, \:and\: solving\: the \:equation \:we\: get \\r= \sqrt{3}

Hence numbers are 1,3,9,27

 

Posted by

Devendra Khairwa

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