The value of the product 6^ (1)/(2) × 6^ (1)/(4) × 6^ (1)/(8) × 6^ (1)/(16) × . up to infinite terms is

Correct answer: A. 6 6^ (1)/(2) × 6^ (1)/(4) × 6^ (1)/(8) × 6^ (1)/(16) × upto infinite terms. =6^ (1)/(2)+(1)/(4)+(1)/(8)+(1)/(16)+ ∞ =6 Sum of ∞ terms of G.P. is S_ ∞ = a 1-r . Here a=(1)/(2), r=(1)/(2) 6^ (1)/(2)+(1)/(4)+(1)/(8)+(1)/(16)+ ∞ =6^ ((1)/(2))/(1-(1)/(2)) =6^ ((1)/(2))/((1)/(2)) =6^1=6 .

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Sequences and Series

NDA 2017 · Mathematics · Sequences and Series

The value of the product 6^ (1)/(2) × 6^ (1)/(4) × 6^ (1)/(8) × 6^...

The value of the product

6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots

. up to infinite terms is

Options

A $6$
B $36$
C $216$
D $512$

Correct answer

A. $6$

Step-by-step solution

6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots

upto infinite terms.

=6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty}=6

Sum of

\infty

terms of G.P. is

\mathrm{S}_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}}

. Here

\mathrm{a}=\frac{1}{2}, \mathrm{r}=\frac{1}{2}\therefore 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty}=6^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=6^{\frac{\frac{1}{2}}{\frac{1}{2}}}=6^1=6

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