Correct Answer - Option 1 : 126
Concept:
Let us consider sequence a1, a2, a3 …. an is an G.P.
nth term of the G.P. is an = arn − 1
Sum of n terms, s = \(\rm \frac{a(r^n-1)}{(r- 1)}\); where r > 1
Sum of n terms, s = \(\rm \frac{a(1-r^n)}{(1-r)}\); where r < 1
Calculation:
The first term for the given series is 2
The second term is 22 = 4,
Hence, the common ratio, r = \(4\over2\) =2
Since, r = 2 > 1
Sum of first 6 terms is \(\rm \frac{a(r^n-1)}{(r- 1)}\)
s = \(\rm \frac{2(2^6-1)}{(2- 1)}\)
⇒ s = 2(64 - 1)
⇒ s = 126