a sum to infinity which is three times the first term. find the first term

Answer 1

a/(1-r) = 3a
divide by a
1/(1-r) = 3
1 = 3 - 3r
3r = 2
r = 2/3
plug in r = 2/3

a/(1 - 2/3) = 3a
a/(1/3) = 3a
3a = 3a , which is a true statement,

so "a" can be anything.
But in a GP, the first term cannot be zero

so the first term can be any real number except zero.