Which term of the G.P 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\) … is \(

1.	Which term of the G.P 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\) … is \(\frac{1}{1024}\)?
Answer» Given, G.P. is 2,1, \(\frac{1}{2}\) , \(\frac{1}{4}\) … First term a = 2 and common ratio = \(\frac{1}{2}\) Let n^th term of G.P. be \(\frac{1}{1024}\) ∴ T_n = arⁿ⁻¹ ⇒ \(\frac{1}{1024}\) = 2(\(\frac{1}{2}\))ⁿ⁻¹ ⇒ \(\frac{1}{2048}\) = (\(\frac{1}{2}\))ⁿ⁻¹ ⇒ (\(\frac{1}{2}\))¹¹ = (\(\frac{1}{2}\))ⁿ⁻¹ ∴ n − 1 = 11 Or n = 12

Which term of the G.P 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\) … is \(\frac{1}{1024}\)?

Answer»

Given,

G.P. is 2,1, \(\frac{1}{2}\) , \(\frac{1}{4}\) …

First term a = 2

and common ratio = \(\frac{1}{2}\)

Let n^th term of G.P. be \(\frac{1}{1024}\)

∴ T_n = arⁿ⁻¹

⇒ \(\frac{1}{1024}\) = 2(\(\frac{1}{2}\))ⁿ⁻¹

⇒ \(\frac{1}{2048}\) = (\(\frac{1}{2}\))ⁿ⁻¹

⇒ (\(\frac{1}{2}\))¹¹ = (\(\frac{1}{2}\))ⁿ⁻¹

∴ n − 1 = 11

Or n = 12