Question

Arrange the geometric series from least to greatest based on the value of their sums.

Answer 1

first is the one with 4 and 2(3) then 5 and 3(2) then 5 and 3 then 7 and 2

Explanation:

Answer 2

80 < 93 < 121 < 127

Explanation:

For a geometric series,

`\sum_(t=1)^(n)a(r)^(t-1)`

Formula to be used,

Sum of t terms of a geometric series =
`(a(r^t-1))/(r-1)`

Here t = number of terms

a = first term

r = common ratio

1).
`\sum_(t=1)^(5)3(2)^(t-1)`

First term of this series 'a' = 3

Common ratio 'r' = 2

Number of terms 't' = 5

Therefore, sum of 5 terms of the series =
`(3(2^5-1))/((2-1))`

= 93

2).
`\sum_(t=1)^(7)(2)^(t-1)`

First term 'a' = 1

Common ratio 'r' = 2

Number of terms 't' = 7

Sum of 7 terms of this series =
`(1(2^7-1))/((2-1))`

= 127

3).
`\sum_(t=1)^(5)(3)^(t-1)`

First term 'a' = 1

Common ratio 'r' = 3

Number of terms 't' = 5

Therefore, sum of 5 terms =
`(1(3^5-1))/(3-1)`

= 121

4).
`\sum_(t=1)^(4)2(3)^(t-1)`

First term 'a' = 2

Common ratio 'r' = 3

Number of terms 't' = 4

Therefore, sum of 4 terms of the series =
`(2(3^4-1))/(3-1)`

= 80

80 < 93 < 121 < 127 will be the answer.