IPMAT IndoreAlgebra > Medium79\frac{7}{9}9723\frac{2}{3}3249\frac{4}{9}9413\frac{1}{3}31✅ Correct Option: 2Related questions:IPMAT Indore 2019There are numbers a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_na1,a2,a3,…,an each of them being +1+1+1 or −1-1−1. 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 = 0a1a2+a2a3+a3a4+…an−1an+ana1=0 thenIPMAT 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 2025Let S1={100,105,110,115,...}S_1 = \{100, 105, 110, 115, ... \}S1={100,105,110,115,...} and S2={100,95,90,85,...}S_2 = \{100, 95, 90, 85, ... \}S2={100,95,90,85,...} be two series in arithmetic progression. If aka_kak and bkb_kbk are the kkk-th terms of S1S_1S1 and S2S_2S2, respectively, then ∑k=120akbk\sum_{k=1}^{20} a_k b_k∑k=120akbk equals __________.