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

Which term of the following sequences is given?

(a) 2, \(2\sqrt{2}\), 4, … is 128 ?

This is a Geometric Progression (G.P.) with \( a = 2,\; r = \sqrt{2} \)

General term of a G.P.: \( T_n = ar^{n-1} \)

\( 2(\sqrt{2})^{\,n-1} = 128 \)

\( (\sqrt{2})^{\,n-1} = 64 = 2^6 \)

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

\( n = 13 \)

Answer: 128 is the 13^th term.

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

This is a G.P. with \( a = \sqrt{3},\; r = \sqrt{3} \)

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

\( (\sqrt{3})^n = 729 \)

\( 729 = 3^6 \Rightarrow (\sqrt{3})^n = 3^6 \)

\( 3^{n/2} = 3^6 \Rightarrow n = 12 \)

Answer: 729 is the 12^th term.

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

This is a G.P. with \( a = \frac{1}{3},\; r = \frac{1}{3} \)

\( \frac{1}{3}\left(\frac{1}{3}\right)^{n-1} = \frac{1}{19683} \)

\( \left(\frac{1}{3}\right)^n = \frac{1}{19683} \)

\( 19683 = 3^9 \Rightarrow \left(\frac{1}{3}\right)^n = \frac{1}{3^9} \)

\( n = 9 \)

Answer: \( \frac{1}{19683} \) is the 9^th term.