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Mathematics

Find the seventh term of the G.P. :

1, 3,3,33,........\sqrt{3}, 3, 3\sqrt{3}, ……..

GP

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Answer

Since,

31=333=3\dfrac{\sqrt{3}}{1} = \dfrac{3\sqrt{3}}{3} = \sqrt{3}

Hence, the above sequence is a G.P. with r = 3\sqrt{3} and a = 1.

We know that nth term of G.P.,

an = arn - 1

a7 = 1.(3)7−1(\sqrt{3})^{7 - 1}

= (3)6(\sqrt{3})^6

= 27.

Hence, 7th term of the G.P. = 27.

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