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