Let a_{1},a_{2},,a_{21} be an AP such that \sum_{n = 1}^{20}\mspace{2mu}\frac{1}{a_{n}a_{n + 1}} = \frac{4}{9}. If the sum of this AP is 189 , then a_{6}a_{16} is equal to: Sequence & Series (Mathematics) | DivineJEE
Exam
Mathematics
Let a1,a2,…,a21 be an AP such that\n ∑n=120\mspace2muanan+11=94.\n If the sum of this AP is 189 , then a6a16 is equal to
A
57
B
48
C
36
D
72
Select an option to instantly check whether it is correct or wrong.
Download Divine JEE App
Learn on the go with our mobile app. Access courses, live classes, and study materials anytime, anywhere.
Empowering students to achieve their IIT dreams through personalized learning and expert guidance.