Checking for common ratio, r
T2/T1 = 1/(1.02)^3/1/(1.02)
= 1/(1.02)^2
Similarly, T3/T2 = 1/(1.02)^5/1/(1.02)^3
= 1/(1.02)^2
Since r is same, the given series is a geometric progression.
With a I.e. first term = 1/1.02
And r I.e. common ratio = 1/(1.02)^2
Sum of infinite GP = a/(1-r)
= (1/1.02)/(1-1/1.0404)
=0.980392156/0.038831218
= 25.25 (approx.)
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