Given: #S_n = 4860;" " r = 3;" " a_n = 3280.5#
#a_n = a_1 r^(n-1); " "S_n = a_1((1-r^n)/(1-r))#
Input what we know into the two equations:
# 3280.5 = a_1 3^(n-1) " " 4860 = a_1((1-3^n)/-2)#
We have two equations and two unknowns.
#a_1 = 3280.5/3^(n-1)#
#4860 = 3280.5/3^(n-1) ((1-3^n)/-2)#
#4860 = 3280.5/-2 ((1-3^n) 3^-(n-1))#
#4860 = -1640.25 (1-3^n) 3^(1-n)#
#4860 = -1640.25 * (3^(1-n) -3^(n+1-n))#
#4860 = -1640.25 * (3^(1-n) -3)#
#4860 = -1640.25 * (3^(1-n)) +4920.75#
#4860 - 4920.75 = -1640.25 * (3^(1-n))#
#-60.75 = -1640.25 * (3^(1-n))#
#(-60.75)/(-1640.25) = 3^(1-n)#
#1/27 = 3^(1-n)#
#3^-3 = 3^(1-n)#
Since both have the same base of #3#, we can set the exponents equal:
#-3 = 1-n#
#-4 = -n " "=> n = 4#
Solve for #a_1#:
#a_1 = 3280.5/3^3 = 121.5#
