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How many terms of G.P. 3, 3^2, 3^3, … are needed to give the sum 120?

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+4 votes
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asked Jan 19, 2017 in Mathematics by Rohit Singh (61,782 points) ● 35 ● 133 ● 355

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+4 votes
answered Jan 20, 2017 by vikash (21,277 points) ● 4 ● 19 ● 70
selected Jan 20, 2017 by Rohit Singh
 
Best answer

Solution:

The given G.P. is 3, 32, 33, …

Let n terms of this G.P. be required to obtain the sum as 120.

S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus space 1 close parentheses over denominator r minus 1 end fraction

Here, a = 3 and r = 3

∴ S subscript n space equals space 120 space equals space fraction numerator 3 space open parentheses 3 to the power of n space minus 1 close parentheses over denominator 3 minus 1 end fractionrightwards double arrow 120 space equals space fraction numerator 3 open parentheses 3 to the power of n space minus 1 close parentheses over denominator 2 end fractionrightwards double arrow fraction numerator 120 space cross times 2 over denominator 3 end fraction space equals space 3 to the power of n space minus 1rightwards double arrow 3 to the power of n space minus 1 space equals space 80rightwards double arrow 3 to the power of n space equals space 81rightwards double arrow 3 to the power of n space equals space 3 to the power of 4

∴  n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.

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