Question 13

How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?

Answer

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

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