Answer:
80 < 93 < 121 < 127
Step-by-step explanation:
For a geometric series,
[tex]\sum_{t=1}^{n}a(r)^{t-1}[/tex]
Formula to be used,
Sum of t terms of a geometric series = [tex]\frac{a(r^t-1)}{r-1}[/tex]
Here t = number of terms
a = first term
r = common ratio
1). [tex]\sum_{t=1}^{5}3(2)^{t-1}[/tex]
First term of this series 'a' = 3
Common ratio 'r' = 2
Number of terms 't' = 5
Therefore, sum of 5 terms of the series = [tex]\frac{3(2^5-1)}{(2-1)}[/tex]
= 93
2). [tex]\sum_{t=1}^{7}(2)^{t-1}[/tex]
First term 'a' = 1
Common ratio 'r' = 2
Number of terms 't' = 7
Sum of 7 terms of this series = [tex]\frac{1(2^7-1)}{(2-1)}[/tex]
= 127
3). [tex]\sum_{t=1}^{5}(3)^{t-1}[/tex]
First term 'a' = 1
Common ratio 'r' = 3
Number of terms 't' = 5
Therefore, sum of 5 terms = [tex]\frac{1(3^5-1)}{3-1}[/tex]
= 121
4). [tex]\sum_{t=1}^{4}2(3)^{t-1}[/tex]
First term 'a' = 2
Common ratio 'r' = 3
Number of terms 't' = 4
Therefore, sum of 4 terms of the series = [tex]\frac{2(3^4-1)}{3-1}[/tex]
= 80
80 < 93 < 121 < 127 will be the answer.
