Question

The sum of the geometric series \(\sqrt{3}+\sqrt{12}+\sqrt{48}+\dots\) up to \(10\) terms is

• A. \(1023\sqrt{3}\)
• B. \(1024\sqrt{3}\)
• C. \(511\sqrt{3}\)
• D. \(512\sqrt{3}\)
• E. \(215\sqrt{3}\)

Hint:

Whenever radicals appear in a sequence, first simplify them. That often reveals a standard A.P. or G.P. pattern and makes the sum much easier to calculate.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem asks for the sum of the first 10 terms of a series. We first need to identify if it is a geometric series by checking for a common ratio. This involves simplifying the terms of the series.
Step 2: Key Formula or Approach:
The sum of the first \(n\) terms of a geometric series is given by the formula:
\[ S_n = \frac{a(r^n - 1)}{r-1} \] where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.
Step 3: Detailed Explanation:
First, simplify the terms of the series:
1st term: \(a_1 = \sqrt{3}\)
2nd term: \(a_2 = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
3rd term: \(a_3 = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\)
The series is \(\sqrt{3} + 2\sqrt{3} + 4\sqrt{3} + ...\)
Next, identify the parameters of the GP:
The first term is \(a = \sqrt{3}\).
The common ratio is \(r = \frac{a_2}{a_1} = \frac{2\sqrt{3}}{\sqrt{3}} = 2\).
(Check: \(\frac{a_3}{a_2} = \frac{4\sqrt{3}}{2\sqrt{3}} = 2\)).
The number of terms is \(n = 10\).
Now, apply the sum formula:
\[ S_{10} = \frac{\sqrt{3}(2^{10} - 1)}{2-1} \] We know that \(2^{10} = 1024\).
\[ S_{10} = \frac{\sqrt{3}(1024 - 1)}{1} \] \[ S_{10} = \sqrt{3}(1023) \] \[ S_{10} = 1023\sqrt{3} \] Step 4: Final Answer:
The sum of the series up to 10 terms is \(1023\sqrt{3}\). Therefore, option (A) is the correct answer.