Question

Let \( b_1 = 3, b_2, b_3, \ldots \) be a geometric progression of increasing positive numbers. Let \[ \sum_{n=1}^{20} b_{3n} = 4\sum_{n=1}^{20} b_{3n-2}. \] Then, the sum of the first ten terms of the G.P. is:

• A. \(1023\)
• B. \(2046\)
• C. \(3069\)
• D. \(3149\)

Hint:

When sums of selected terms of a G.P. are related, factor out common powers of the ratio to simplify the equation.

Answer 1

The Correct Option is C

Concept: For a geometric progression:
The \(n^{\text{th}}\) term is \( b_n = ar^{\,n-1} \)
The sum of the first \(n\) terms is \[ S_n = a\frac{r^n - 1}{r - 1}, \quad r \neq 1 \]
Given summation conditions can be used to determine the common ratio \(r\).
Step 1: Write the general term Given \( b_1 = 3 \), let the common ratio be \(r\). Then: \[ b_n = 3r^{n-1} \]
Step 2: Express the given sums \[ \sum_{n=1}^{20} b_{3n} = \sum_{n=1}^{20} 3r^{3n-1} \] \[ \sum_{n=1}^{20} b_{3n-2} = \sum_{n=1}^{20} 3r^{3n-3} \]
Step 3: Use the given condition \[ \sum_{n=1}^{20} 3r^{3n-1} = 4\sum_{n=1}^{20} 3r^{3n-3} \] Cancel \(3\) from both sides: \[ \sum_{n=1}^{20} r^{3n-1} = 4\sum_{n=1}^{20} r^{3n-3} \] Factor out \( r^2 \) from the left: \[ r^2 \sum_{n=1}^{20} r^{3n-3} = 4\sum_{n=1}^{20} r^{3n-3} \] Since the sum is nonzero: \[ r^2 = 4 \Rightarrow r = 2 \quad (\text{as the G.P. is increasing}) \]
Step 4: Find the sum of the first ten terms \[ S_{10} = 3\frac{2^{10} - 1}{2 - 1} = 3(1024 - 1) = 3069 \]