So we have:
#a + ar+ar^2+ar^3+ar^4+ ...=13.5#
and
#a+ ar+ar^2 = 13#
Subtracting we have:
#(a + ar+ar^2+ar^3+ar^4+ ...) - (a+ ar+ar^2) = 0.5#, that is:
#ar^3 + ar^4+ar^5+ar^6+ar^7+ ...=0.5#
Hence, taking #r# as a common factor:
# r^3 * (a + ar+ar^2+ar^3+ar^4+ ...)=0.5#
But we know that that the expression between brackets:
#a + ar+ar^2+ar^3+ar^4+ ...=13.5#
so we have:
# r^3 * 13.5=0.5#, that is:
# r^3 =0.5/13.5#, that is:
# r^3 =(1/2)/(27/2)=1/27#, and then
#r=1/3#
But we know that #a+ ar+ar^2 = 13#, so:
#a+ a1/3+a1/9 = 13#, and that means:
#(9a+3a+a)/9=13#, so:
#(13a)/9=13#, so #a/9=1#, and from that we get:
#a=9#, which is the first term
