For a geometric series:
#S_n=(a(1-r^n))/(1-r)#
Where:
#bba# is the first term, #bbr# is the common ratio and #bbn# is the nth term.
We know the first term #a_1=3# and common ratio is #-5#
We need to find the value of #n#.
If we find the sum of the first #n# terms and then subtract the sum of the first #n-1# terms, this will give us the #a_n#th term.
#(3(1+5^(n-1)))/(1+5)=(1+(5)^(n-1))/2#
#(1+(5)^(n))/2-(1+(5)^(n-1))/2=a_n=46875#
#(1+(5)^(n)-1-(5)^(n-1))/2=46875#
#((5)^(n)-(5)^(n-1))/2=46875#
#(5)^(n)-(5)^(n-1)=93750#
#5^n=5^(n-1+1)#
#5^(n-1+1)=5^1*5^(n-1)#
#5^1*5^(n-1)-5^(n-1)=93750#
#5^(n-1)(5^1-1)=93750#
#5^(n-1)*4=93750#
#5^(n-1)=93750/4=23437.5#
Taking logs:
#(n-1)ln(5)=ln(23437.5)#
#n=ln(23437.5)/ln(5)+1~~7.251929636#
#n# must be an integer so #n=7#
#S_n=S_7=(3(1+5^7))/6=234378/6=39063#