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)
