Question

Which term of the following sequences: 

(a) 2 ,\(2\sqrt{2}\) , 4 ,.... is 128 ?

 (b) \(\sqrt{3}\), 3, \(3\sqrt{3}\),... is 729 ? 

(c) \(\frac{1}{3},\frac{1}{9},\frac{1}{27}\) ,.... is \(\frac{1}{19683}\) ?

Answer 1

(a) The given sequence is 2 , \(2\sqrt{2}\) , 4,....

Here, a = 2 and r = \(\frac{2\sqrt 2}{2}\)= \(\sqrt{2}\)

Let the n th term of the given sequence be 128

an = ar n - 1

⇒ (2)\((\sqrt{2})\) n - 1 = 128

⇒ (2)(2)n - \(\frac{1}{2}\) = (2\()^7\)

⇒ (2)n - \(\frac{1}{2}\) + 1 = (2\()^7\)

⇒ ∴ n - \(\frac{1}{2}\) + 1 = 7

⇒ n - \(\frac{1}{2}\) = 6

⇒ n - 1 = 12

⇒ n = 13

Thus, the 13th term of the given sequence is 128.

(b) The given sequence is \(\sqrt{3}\) , 3 , \(3\sqrt{3}\) ,...

Here, a = \(\sqrt{3}\) and r = \(\frac{3}{\sqrt{3}}\)= \(\sqrt{3}\)

Let the n th term of the given sequence be 729.

an = arn - 1

∴ arn - 1 = 729

⇒ (\(\sqrt{3}\))(\(\sqrt{3}\)n - 1) = 729

⇒ (3)\(\frac{1}{2}\) (3)n - \(\frac{1}{2}\) = (3)\(^6\)

⇒ (3)\(\frac{1}{2}\) + n - \(\frac{1}{2}\) = (3\()^6\)

∴ \(\frac{1}{2}\) + n - \(\frac{1}{2}\)= 6

⇒ n = 12

Thus, the 12th term of the given sequence is 729.

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(c) The given sequence is \(\frac{1}{3},\frac{1}{9},\frac{1}{12}\),…..

Here, a = \(\frac{1}{3}\) and r = \(\frac{1}{9}\) ÷ \(\frac{1}{3}=\frac{1}{3}\)

Let the n th term of the given sequence be \(\frac{1}{19683}\)

an = arn \(^-1\)

∴ arn \(^-1\) = \(\frac{1}{19683}\)

⇒ \((\frac{1}{3})(\frac{1}{3})\)n \(^6\) = \(\frac{1}{19683}\)

⇒\((\frac{1}{3})\)n = \((\frac{1}{3})^9\)

⇒ n = 9

Thus, the 9th term of the given sequence is \(\frac{1}{19683}\)