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

Answer:80 < 93 < 121 < 127Step-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 termsa = first termr = common ratio1). [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]                                                                       = 932). [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]                                                     = 1273). [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]                                                  = 1214). [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.