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IPMAT Indore 2025 (MCQ) PYQs

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IPMAT Indore 2025

Algebra
>
Progression & Series

Medium

Given that 1+122+132+142+...=π261 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}, the value of 1+132+152+172+...1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ... is

✅ Correct Option: 4

Note: This IPMAT Indore 2025 question is repeated from IPMAT Indore 2020.

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There are numbers a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n each of them being +1+1 or −1-1. If it is known that a1a2+a2a3+a3a4+…an−1an+ana1=0a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots a_{n-1} a_n + a_n a_1 = 0 then

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