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CAT 2024 Slot 2 — QA Question 6

Infinite Geometric ProgressionEasy

The sum of the infinite series 15(15-17) + (15)2((15)2-(17)2) + (15)3((15)3-(17)3) + ... is equal to

Answer & solution

  • 5/408

  • B

    5/816

  • C

    7/408

  • D

    7/816

Solution

Medium

The nnth term is (15)n ⁣[(15)n−(17)n]\left(\tfrac15\right)^n\!\left[\left(\tfrac15\right)^n-\left(\tfrac17\right)^n\right]. Expand it into two geometric series with ratios 125\tfrac{1}{25} and 135\tfrac{1}{35} and sum each.

1

Write the general term. Multiplying out:

tn=(15) ⁣n ⁣[(15) ⁣n−(17) ⁣n]⇒ tn=(125) ⁣n−(135) ⁣n\begin{aligned} &t_n=\left(\tfrac15\right)^{\!n}\!\left[\left(\tfrac15\right)^{\!n}-\left(\tfrac17\right)^{\!n}\right]\\ &\Rightarrow\ t_n=\left(\tfrac{1}{25}\right)^{\!n}-\left(\tfrac{1}{35}\right)^{\!n} \end{aligned}
2

Sum the two infinite GPs. Each starts at n=1n=1, so ∑n≥1rn=r1−r\sum_{n\ge1}r^n=\dfrac{r}{1-r}.

S=∑n≥1(125) ⁣n−∑n≥1(135) ⁣n⇒ S=1251−125−1351−135(GP sum formula)⇒ S=124−134\begin{aligned} &S=\sum_{n\ge1}\left(\tfrac{1}{25}\right)^{\!n}-\sum_{n\ge1}\left(\tfrac{1}{35}\right)^{\!n}\\ &\Rightarrow\ S=\frac{\tfrac{1}{25}}{1-\tfrac{1}{25}}-\frac{\tfrac{1}{35}}{1-\tfrac{1}{35}} \quad\text{(GP sum formula)}\\ &\Rightarrow\ S=\frac{1}{24}-\frac{1}{34} \end{aligned}
3

Combine.

S=34−2424⋅34=10816=5408\begin{aligned} &S=\frac{34-24}{24\cdot 34}=\frac{10}{816}=\frac{5}{408} \end{aligned}
S=5408S=\dfrac{5}{408}
CAT 2024 Slot 2 QA Q6: The sum of the infinite series 1 5 1 5 - 1 7 + 1 5 2 1 5 2 - 1 7 2 + 1 5 3 1 5 3 - 1 7 3 + ... is equal to — Solution | TheCATExam