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Permutations And Combinations

Question
CBSEENMA11013091
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Given a G.P. with a = 729 and 7th term 64, determine S7.

Solution

Here a = 729, Let r be the common ratio
space space straight t subscript 7 space equals space 64 space space space space space space space space space space space space space space space rightwards double arrow space space space ar to the power of 7 minus 1 end exponent space equals space 64 space rightwards double arrow space 729. space straight r to the power of 6 space equals space 64
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When space space straight r space equals space 2 over 3 comma  then
 straight S subscript 7 space equals space straight a fraction numerator left parenthesis 1 minus straight r to the power of 7 right parenthesis over denominator 1 minus straight r end fraction 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 equals space 2187 open square brackets 1 minus 128 over 2187 close square brackets space equals space 2187 space open square brackets fraction numerator 2187 minus 128 over denominator 2187 end fraction close square brackets space equals space 2059
When straight r space equals space fraction numerator negative 2 over denominator 3 end fraction comma  then
straight S subscript 7 space equals space fraction numerator straight a left parenthesis 1 minus straight r to the power of 7 right parenthesis over denominator 1 minus straight r end fraction space equals space fraction numerator 729 open square brackets 1 minus open parentheses begin display style fraction numerator negative 2 over denominator 3 end fraction end style close parentheses to the power of 7 close square brackets over denominator 1 minus open parentheses begin display style fraction numerator negative 2 over denominator 3 end fraction end style close parentheses end fraction space equals space 2187 over 5 open square brackets 1 plus 128 over 2187 close square brackets space equals space fraction numerator 2187 plus 128 over denominator 5 end fraction space equals space 2315 over 5 space equals space 463

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