Question

The sum of the first 20 terms of the G.P. \( \sqrt{3} + \frac{-1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \cdots \) is equal to

• A. \( \frac{\sqrt{3}(3^{20} - 1)}{4 \cdot 3^{19}} \)
• B. \( \frac{\sqrt{3}(3^{20} - 1)}{2 \cdot 3^{19}} \)
• C. \( \frac{\sqrt{3}(3^{20} - 1)}{3^{20}} \)
• D. \( \frac{\sqrt{3}(3^{20} - 1)}{3^{20}} \)
• E. \( \frac{\sqrt{3}(3^{20} - 1)}{2 \cdot 3^{20}} \)

Hint:

For G.P. sum, use the formula \( S_n = \frac{a(1 - r^n)}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio.

Answer 1

The Correct Option is A

This is a geometric progression (G.P.) where the first term \( a = \sqrt{3} \), and the common ratio \( r = \frac{-1}{\sqrt{3}} \). The sum of the first \( n \) terms of a G.P. is given by: \[ S_n = \frac{a(1 - r^n)}{1 - r} \] Substituting the values for \( a \), \( r \), and \( n = 20 \): \[ S_{20} = \frac{\sqrt{3}(1 - \left( \frac{-1}{\sqrt{3}} \right)^{20})}{1 - \left( \frac{-1}{\sqrt{3}} \right)} \] \[ S_{20} = \frac{\sqrt{3}(1 - \frac{1}{3^{10}})}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}(3^{20} - 1)}{4 \cdot 3^{19}} \] Thus, the correct answer is (A).