IPMAT Indore 2025Algebra > 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:The terms of a geometric progression are real and positive. If the ppp-th term of the progression is qqq and the qqq-th term is ppp, then the logarithm of the first term isIf 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∞ isThe 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 is