Prove that :
(91/3 . 91/9 . 91/27 ….∞) = 3.
Using the properties of exponents:
The above term can be written as
Let S = …(1)
We observe that above progression(in power of 9) possess a common ratio. So it is a geometric progression.
Let m =
Common ratio = r =
Sum of infinite GP = ,where a is the first term and r is the common ratio.
Note: We can only use the above formula if |r|<1
Clearly, a = and r =
⇒ m =
From equation 1 we have,
S = 9m = 91/2 = 3 = RHS
Hence Proved