Arrange the geometric series from least to greatest based on the value of their sums
Accepted Solution
A:
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.