Which term of the GP√3, 3, 3√3… is 729?

1.	Which term of the GP√3, 3, 3√3… is 729?
Answer» Given GP is √3, 3, 3√3…. The given GP is of the form, a, ar, ar² , ar³…. Where r is the common ratio. First term in the given GP, a₁ = a = √3 Second term in GP, a₂ = 3 Now, the common ratio, \(r = \frac{a_2}{a_1}\) r = \(\frac{3}{√3}\) = √3 Let us consider 729 as the n^th term of the GP. Now, nth term of GP is, aⁿ = ar^{n – 1} 729 = √3 (√3)^{n – 1} √3ⁿ = √3¹² n = 12 So, 729 is the 12^th term in GP.

Answer»

Given GP is √3, 3, 3√3….

The given GP is of the form, a, ar, ar² , ar³….

Where r is the common ratio.

First term in the given GP,

a₁ = a = √3

Second term in GP, a₂ = 3

Now, the common ratio, \(r = \frac{a_2}{a_1}\)

r = \(\frac{3}{√3}\) = √3

Let us consider 729 as the n^th term of the GP. Now, nth term of GP is, aⁿ = ar^{n – 1}

729 = √3 (√3)^{n – 1}

√3ⁿ = √3¹²

n = 12

So, 729 is the 12^th term in GP.

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