−1/3
2
4/3
2/9
Correct answer is A
T2=−23;S∞32
Tn=arn−1
∴ T2=ar=−23---eqn.(i)
S∞=a1−r=32---eqn.(ii)
= 2a = 3(1 - r)
= 2a = 3 - 3r
∴ a = 3−3r2
Substitute 3−3r2 for a in eqn.(i)
= 3−3r2×r=−23
= 3r−3r22=−23
= 3(3r - 3r2) = -4
= 9r - 9r2 = -4
= 9r2 - 9r - 4 = 0
= 9r2 - 12r + 3r - 4 = 0
= 3r(3r - 4) + 1(3r - 4) = 0
= (3r - 4)(3r + 1) = 0
∴ r = 43or−13
For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.
∴ r = -1/3 (since |-1/3| < 1)
What is the mode of the numbers 8, 10, 9, 9, 10, 8, 11, 8, 10, 9, 8 and 14? ...
In the diagram PQRS is a circle, |PT| = |QT| and ∠QPT = 70o what is the size of ∠PRS?...
Given that loga2 = 0.693 and loga3 = 1.097, find loga 13.5...
What is the locus of point that is equidistant from points P(1,3) and Q(3,5)? ...
The bearing of P from Q is x, where 270o < x < 360o. Find the bearing of Q from P...