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+1 vote
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in Geometric Progressions by (15.7k points)

The 4th and 7th terms of a GP are \(\frac{1}{27}\) and \(\frac{1}{729}\) respectively. Find the sum of n terms of the GP.

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1 Answer

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4th term = ar4-1 = ar3\(\frac{1}{27}\) 

7th term = ar7-1= ar6\(\frac{1}{729}\)

Dividing the 7th term by the 4th term

\(\frac{ar^6}{ar^3} = \cfrac{\frac{1}{729}}{\frac{1}{27}}\)

⇒ r3 = \(\frac{1}{27}\) ........... (1)

∴ r = \(\frac{1}{3}\)

ar3 = \(\frac{1}{27}\) [Putting from eqn (i)]

a\(\frac{1}{27}\) = \(\frac{1}{27}\)

∴ a = 1

Sum of a G.P. series is represented by the formula, Sn = \(a\frac{1 - r^n}{1-r}\), when |r|<1. 

‘Sn’ represents the sum of the G.P. series up to nth terms, 

‘a’ represents the first term, 

‘r’ represents the common ratio and 

‘n’ represents the number of terms. 

Here,

a = 1 

r = \(\frac{1}{3}\)

n terms

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