IPMAT Indore 2024Algebra > Hard(1−q)log(p)−(1−p)log(q)p−q\dfrac{(1-q)\log(p)-(1-p)\log(q)}{p-q}p−q(1−q)log(p)−(1−p)log(q)(1−q)log(q)−(1−p)log(p)p−q\dfrac{(1-q)\log(q)-(1-p)\log(p)}{p-q}p−q(1−q)log(q)−(1−p)log(p)(1−q)log(p)+(1−p)log(q)p−q\dfrac{(1-q)\log(p)+(1-p)\log(q)}{p-q}p−q(1−q)log(p)+(1−p)log(q)(1−q)log(q)+(1−p)log(p)p−q\dfrac{(1-q)\log(q)+(1-p)\log(p)}{p-q}p−q(1−q)log(q)+(1−p)log(p)✅ Correct Option: 2Related questions:IPMAT Indore 2023A person standing at the centre of an open ground first walks 32 meters towards the east, takes a right turn and walks 16 meters, takes another right turn and walks 8 meters, and so on. How far will the person be from the original starting point after an infinite number of such walks in this pattern?IPMAT Indore 2024The sum of a given infinite geometric progression is 80 and the sum of its first two terms is 35. Then the value of nnn for which the sum of its first nnn terms is closest to 100, 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 __________.