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CAT 2024 Slot 2 QA Question 10

Multiple choice (+3 / −1) · Algebra · Progression & Series · Try it, then check the answer and solution below.

CAT 2024 Slot 2QAAlgebra • Progression & SeriesModerate
The sum of the infinite series is 15\frac{1}{5} \left( 15\frac{1}{5} - 17\frac{1}{7} \right)+\left( 15\frac{1}{5} \right)^{2}\left(\left( 15\frac{1}{5} \right)^{2}-\left( 17\frac{1}{7} \right)^{2}\right)+\left( 15\frac{1}{5} \right)^{3}\left(\left( 15\frac{1}{5} \right)^{3}-\left( 17\frac{1}{7} \right)^{3}\right)+\ldots$ equal to
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Answer and solution

Answer: A) 5408\frac{5}{408}

📌 Core Concept & Formula
The governing mathematical theorem for this problem is the sum of an infinite geometric series. The formula for the sum of an infinite geometric series starting from n=1n = 1 is:
Sum=a1−r\text{Sum} = \frac{a}{1 - r}
where aa is the first term and rr is the common ratio, provided that ∣r∣<1|r| < 1.
🔢 Step-by-Step Derivation
1
Identify the General Term:
The given series is:
∑n=1∞((15)n((15)n−(17)n))\sum_{n=1}^{\infty} \left( \left( \frac{1}{5} \right)^n \left( \left( \frac{1}{5} \right)^n - \left( \frac{1}{7} \right)^n \right) \right)
Simplifying the general term:
(15)n((15)n−(17)n)=(125)n−(135)n\left( \frac{1}{5} \right)^n \left( \left( \frac{1}{5} \right)^n - \left( \frac{1}{7} \right)^n \right) = \left( \frac{1}{25} \right)^n - \left( \frac{1}{35} \right)^n
2
Split the Series:
The series can be split into two separate geometric series:
∑n=1∞(125)n−∑n=1∞(135)n\sum_{n=1}^{\infty} \left( \frac{1}{25} \right)^n - \sum_{n=1}^{\infty} \left( \frac{1}{35} \right)^n
3
Calculate Each Series:
For the first series with a=125a = \frac{1}{25} and \( r = \frac{1}{2

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