Question

for a sequence if sn=2(3n-1)find the nth term hence show that the sequence is aG.P​

Answer 1

Answer:

The nth term of the sequence is 6.

Step-by-step explanation:

Given : A sequence $S_n=2(3n-1)$

To find : The nth term and hence show that the sequence is a G.P?

Solution :

The sum of sequence is $S_n=2(3n-1)=6n-2$

$a_n=S_n-S_{n-1}$

$a_n=6n-2-(6(n-1)-2)$

$a_n=6n-2-(6n-6-2)$

$a_n=6n-2-(6n-8)$

$a_n=6n-2-6n+8$

$a_n=6$

The nth term of the sequence is 6.

Substitute n=1 in sum of sequence,

$S_1=2(3(1)-1)$

$S_1=2(2)$

$S_1=4$

The first term of sequence is a=4

Substitute n=2 in sum of sequence,

$S_2=2(3(2)-1)$

$S_2=2(5)$

$S_2=10$

i.e. Sum of first two term is 10.

$a_1+a_2=10$

$4+a_2=10$

$a_2=6$

nth term is also 6.

Which means the sequence has only 2 terms

The first term is a=4 and second term is 6.

Answer 2

Step-by-step explanation:

T

n

=S

n

−S

n−1

=2n

2

+5n−[2(n−1)

2

+5(n−1)]

=2n

2

+5n−2n

2

−2+4n−5n+5

∴T

n

=4n+3

T

1

=4(1)+3=7

T

2

=4(2)+3=11

T

3

=4(3)+3=15