Question 15

Given a G.P. with a = 729 and 7th term 64, determine S7.

Answer

a = 729

a7 = 64

Let r be the common ratio of the G.P.

It is known that, an = a rn–1

a7 = ar7–1 = (729)r6

⇒ 64 = 729 r6

rightwards double arrow r to the power of 6 space equals space 64 over 729
rightwards double arrow r to the power of 6 space space equals space open parentheses 2 over 3 close parentheses to the power of 6
rightwards double arrow r space equals space 2 over 3

Also, it is known that, 

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

therefore space S subscript 7 space equals space fraction numerator 729 open square brackets 1 minus open parentheses begin display style 2 over 3 end style close parentheses to the power of 7 close square brackets over denominator 1 minus begin display style 2 over 3 end style end fraction
space space space space space space space space space space space space equals space 3 space cross times space 729 space open square brackets 1 minus open parentheses 2 over 3 close parentheses to the power of 7 close square brackets
space space space space space space space space space space space space equals open parentheses 3 close parentheses to the power of 7 space open square brackets fraction numerator open parentheses 3 close parentheses to the power of 7 space minus space open parentheses 2 close parentheses to the power of 7 over denominator open parentheses 3 close parentheses to the power of 7 end fraction close square brackets
space space space space space space space space space space space space equals space open parentheses 3 close parentheses to the power of 7 space minus open parentheses 2 close parentheses to the power of 7
space space space space space space space space space space space space space equals space 2187 space minus space 128
space space space space space space space space space space space space space equals space 2059

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