We can solve this problem using nth term of a G.P. If a is the first term and r id the common ratio of a G.P. then its nth term tn is given by tn = arn−1 If a is the first term and r is the common ratio of a G.P. then the G.P. can be written as a , ar , ar2 , ... , arn−1 , ... (a ≠0) By using this definition we can find the answer Let G.P. be a,ar,ar2 , ... Then, tp+q = arp+q−1 = m ... (i) tp−q = arp−q−1 = n ... (ii) Dividing (i) by (ii) we get the value of r
arp+q−1
arp−q−1
=
m
n
rp+q−1−(p−q−1) =
m
n
rp+q−1−p+q+1 =
m
n
r2q =
m
n
r = (
m
n
)
1
2q
By substituting r = (
m
n
)
1
2q
in equation (i) we get the value of a arp+q−1 = m a(