IPMAT IndoreAlgebra > Mediumπ26−1\frac{\pi^2}{6} - 16π2−1π6\frac{\pi}{6}6ππ212\frac{\pi^2}{12}12π2π28\frac{\pi^2}{8}8π2✅ Correct Option: 4Related questions:IPMAT Indore 2022The numbers −16,2x+3−22x−1−16,22x−1+16-16,2^{x+3}-2^{2 x-1}-16,2^{2 x-1}+16−16,2x+3−22x−1−16,22x−1+16 are in an arithmetic progression. Then xxx equals ________.IPMAT Indore 2022The sum of the first 15 terms in an arithmetic progression is 200, while the sum of the next 15 terms is 350. Then the common difference isIPMAT Indore 2020If 112+122+132+…\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots121+221+321+… up to ∞=π26\infty = \frac{\pi^2}{6}∞=6π2, then the value of 112+132+152+…\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots121+321+521+… up to ∞\infty∞ is