Question

for a sequence,if Sn= 2(3^n-1)find the n th term hence show that the sequence is G.P.


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Answer 1

Answer:

The sequence is 4,12,36...

is GP.

$a_n = 4( {3}^{n - 1} )$

Step-by-step explanation:

As it is given that sum of n terms

$S_n = 2( {3}^{n} - 1)$

To find the nth term,firstly find the sequence

1. Put n= 1(sum of 1 term = first term)

$a = 2( {3}^{1} - 1) \\ \\ = 2 \times 2 \\ \\ a= 4$

2. Put n= 2

$S_2 = 2( {3}^{2} - 1) \\ \\ = 2 \times (9 - 1) \\ \\ = 2 \times 8 \\ \\ S_2= 16 \\ \\ S_2 = a + a_1 \\ \\ 16 = 4 + a_1 \\ \\ a_1 = 12$

second term is 12

for 3rd term,put n= 3

$S_3 = a + a_1 + a_2 \\ \\ 2 \times ( {3}^{3} - 1) = 16 + a_2 \\ \\ 2 \times 26 = 16 + a_2 \\ \\ 52 - 16 = a_2 \\ \\ a_2 = 36$

Hence the GP is 4,12,36...

nth term of sequence is =

$a {r}^{(n-1)} \\ \\ = 4( {3}^{n - 1} )$

first term a= 4

common ratio r= 3

Hence the sequence s GP.

Hope it helps you.