Question

please answer it guys ​

Answer 1

Answer:

nth term = 4×(3)^n-1

Step-by-step explanation:

Given S n = 2(3^n -1)

for n =1, S1 = first term = t1 =2(3-1) =4

for n= 2 , S2 = the sum o f first two terms = 2(9-1) =16

now t2 = S2 -S1 =16- 4 =12

similarly S3 = 2(26) =52

and  t3 = S3 -S2 =52 -16 = 36 and so on

Now we can write the series as 4,12,36,...

with a common ratio 3. ∴ the sequence is in G.P

now the nth term can be written as a×r^(n-1), where 'r' is  the common ratio and  'a' is the first term .

nth term = 4×(3)^n-1