The fourth term of a GP. is 27 and the 7th term is 729, find the GP.

Solution

Let a be the first term and r be the common ratio
          $space space straight t subscript 4 space equals space 27 space space space rightwards double arrow space space ar to the power of 4 minus 1 end exponent space equals space 27 space rightwards double arrow space ar cubed space equals space 27$                      ...(i)
           $space space straight t subscript 7 space equals space 729 space rightwards double arrow space space ar to the power of 7 minus 1 end exponent space equals space 729 space rightwards double arrow space ar to the power of 6 space equals space 729$                 ...(ii)
Dividing (ii) by (i), we get
                 $space space fraction numerator ar to the power of 6 over denominator ar cubed space end fraction space equals space 729 over 27 space rightwards double arrow space straight r cubed space equals space 27 space rightwards double arrow space space straight r space equals space 3$
Using r = 3 in (i), we get
                           $straight a left parenthesis 3 right parenthesis cubed space equals space 27 space rightwards double arrow space straight a space equals space 27 over 27 space equals space 1$

∴ G.P. is 1, 1(3), 1(3)², 1(3)³, ........... or 1, 3, 9, 27

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Permutations And Combinations

The fourth term of a GP. is 27 and the 7th term is 729, find the GP.

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