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Mathematics

Which term of the G.P. 3,33,93,\sqrt{3}, 3\sqrt{3}, 9\sqrt{3}, \dots is 7293729\sqrt{3} ?

  1. 7th

  2. 6th

  3. 9th

  4. 8th

G.P.

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Answer

In the given A.P.,

a = 3\sqrt3

r = 333\dfrac{3\sqrt3}{\sqrt3} = 3

Let nth term of G.P. be 7293729\sqrt{3}.

⇒ Tn = 7293729\sqrt3

3×3n1=7293\sqrt{3} \times 3^{n - 1} = 729\sqrt{3}

⇒ (3)n - 1 = 729

⇒ (3)n - 1 = 36

⇒ n - 1 = 6

⇒ n = 6 + 1

⇒ n = 7.

Hence, option 1 is the correct option.

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