Permutations and Combinations

Permutations and Combinations

Question

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

Answer

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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