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 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 2025The sum of the first 5 terms of a geometric progression is the same as the sum of the first 7 terms of the same progression. If the sum of the first 9 terms is 24, then the 4th term of the progression isIPMAT 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, is