Find the sum of the 10th and 11th terms of the G.P. The third term of a geometric progression of positive terms is $\frac{6}{25}$ and the seventh term is $1\frac{23}{27}$. Find the sum of the 10th and 11th terms of the G.P., giving your answers correct to 2 decimal places.
Please help, i know to find a and r, i also know to find the sum, but what does 

the sum of 10th and 11th terms 

mean?
P.S $a = \frac{54}{625}$ and $r=1\frac{2}{3}$ please tell me if it correct or no.

the first term is $a$, the second is $ar$ and so on until you hit the $n$th term is $ar^{n-1}$, so the sum of the 10th and 11th term is:
$$ar^{9}+ar^{10}=ar^9(1+r)$$
$$={54\over 625}\cdot \left({5\over 3}\right)^9\left({8\over 3}\right)$$
