n^th term of a G.P series is 2ⁿ, then what is the sum of it's first six terms?

Question

Question

Answer 1

Concept:

Let us consider sequence a₁, a₂, a₃ …. an is an G.P.

n^th term of the G.P. is an = ar^{n − 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