Find the 8th term and the sum of the series 3,6,12,_____ up to 8 terms

Answer 1

a=3
r = 2
a8 = ar^7
S8 = a(r^8 - 1)/(r-1)
So plug and chug

Answer 2

That is geometric progression where:

a1 = initial value

r = common ratio

Common ratio:

r = a2 / a1 = 6 / 3 = 2

r = a3 / a2 = 12 / 6 = 2

In this case a1 = 3 , r = 2

a1 = 3

a2 = a1 ∙ r = 3 ∙ 2 = 6

a3 = a2 ∙ r = 6 ∙ 2 = 12

a4 = a3 ∙ r = 12 ∙ 2 = 24

a5 = a4 ∙ r = 24 ∙ 2 = 48

a6 = a5 ∙ r = 48 ∙ 2 = 96

a7 = a6 ∙ r = 96 ∙ 2 = 192

a8 = a7 ∙ r = 192 ∙ 2 = 384

nth partial sum of a geometric sequence:

Sn = a1 ( 1 - r ⁿ ) / ( 1 - r )

In this case:

S8 = a1 ( 1 - r ⁸ ) / ( 1 - r )

S8 = 3 ∙ ( 1 - 2 ⁸ ) / ( 1 - 2 )

S8 = 3 ∙ ( 1 - 256 ) / ( - 1 )

S8 = 3 ∙ ( - 255 ) / ( - 1 )

S8 = 3 ∙ 255 = 765

You can chect that:

3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765

Answer 3

What is the sum of the first 8 terms of a geometric sequence given it's a¹=2 and r=–4?