Question

The nth term, `U_n`, of a geometric sequence is given by `U_n=3(4)^(n+1)`, a) Find the common ratio `r`, b) Hence, or otherwise, find `S_n`, the sum of the first `n` terms of this sequence?

`U_n=3(4)^(n+1)`

Answer 1

(a) Common ratio is `r=4`, and (b) Sum: `S_n=16(4^n-1)`

Explanation:

`U_n= 3(4)^(n+1) or U_n= 3(4)^2 *4^(n-1)` or

(a) `U_n= 48 (4^(n-1)) (1)` . We know `n` th term of a

geometric sequence with first term `a` and common

ratio `r` is `U_n=a(r)^(n-1) (2)` Comparing both (1) and

(2) we get first term as `a=16` and common ratio as `r=4`.

(b) Sum of this geometric sequence is

`S_n=(a (r^n-1))/(r-1) = (48(4^n-1))/(4-1)=16(4^n-1)` [Ans]