Question

If the sum of an infinite go is 4/3 and the sum of the series obtained on squaring each term is 16/27 then common ratio is

Answer 1

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Answer 2

The common ratio of the given GP is 1/2.

Given:

The Sum of an infinite term in GP is 4/3 and the sum of the series obtained on squaring each term is 16/27.

To Find:

The common ratio

Solution:

Let the first term of the infinite term GP be a and the common ratio be r.

Formula

• Sum Of GP(infinite terms) = $\frac{a}{1-r}$

Given Sum Of GP(infinite terms) = 4/3

$\frac{a}{1-r} = \frac{4}{3\\}\\ \\a=\frac{4(1-r)}{3} -----------------------(1)$

After Squaring each term, a² is the first term and r² is the common ratio.

Sum of GP (After squaring each term ) = 16/17

$\frac{a^{2} }{1-r^{2} } = \frac{16}{27} \\\\a^{2} =\frac{16(1-r^{2)} }{27} --------------------(2)$

Now, Squaring on both sides in equation (1)

$a^{2} = \frac{16(1-r)^{2} }{9} ---------------------(3)$

From Equations  (2) and (3)

$\frac{16(1-r^{2}) }{27} = \frac{16(1-r)^{2} }{9} \\\\ \frac{1-r^{2} }{27} = \frac{(1-r)^{2} }{9} \\\\ \frac{1-r^{2} }{(1-r)^{2} } = \frac{27}{9}\\\\\frac{(1+r)(1-r)}{(1-r)(1-r)} = 3\\\\\\\frac{1+r}{1-r} = 3\\\\1+r=3-3r\\\\4r=2\\r=\frac{2}{4} \\\\r =\frac{1}{2}$

∴ The common ratio of the given GP is 1/2.

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