IPMAT IndoreAlgebra > Hardπ28\frac{\pi^2}{8}8π2π216\frac{\pi^2}{16}16π2π212\frac{\pi^2}{12}12π2π236\frac{\pi^2}{36}36π2✅ Correct Option: 1Related questions:IPMAT Indore 2025Given that 1+122+132+142+...=π261 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}1+221+321+421+...=6π2, the value of 1+132+152+172+...1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ...1+321+521+721+... isIPMAT Indore 2022A new sequence is obtained from the sequence of positive integers (1,2,3,…)(1,2,3, \ldots)(1,2,3,…) by deleting all the perfect squares. Then the 2022nd 2022^{\text {nd }}2022nd term of the new sequence is ________.IPMAT Indore 2019The number of terms common to both the arithmetic progressions 2,5,8,11,...,1792, 5, 8, 11, ..., 1792,5,8,11,...,179 and 3,5,7,9,...,1013, 5, 7, 9, ..., 1013,5,7,9,...,101 is