Geometrical progression is given by the following formula:

a_{n+1}=a_{n}*q, where q is the common ratio. From this formula, you can obtain another one, relating the first term of progression to the arbitrary one,

a_{n}=a_{1}*q^{n-1};

So, for the sixth term in your case, we need to determine q and a_{1}. a_{1}=625; q=125/625, ratio of two successive terms, q=1/5.

a_{6}=625*(1/5)^{5}=625/5^{5}=1/5;

Now, for the sum of N terms, from the first to the N^{th} one.

S_{N}=a_{1}+a_{2}+…+a_{N}=a_{1}+a_{1}*q+a_{1}*q^{2}+…+a_{1}*q^{N-1}=a_{1}*(1+q+…+q^{N-1})

Let us multiply and divide this expression by (q-1). We will obtain the following:

S_{N}=a_{1}*(q-1)*(1+q+…+q^{N-1})/(q-1)=a_{1}*(q^{N}-1)/(q-1)

Thus, for the sum of 15 terms we will get:

S_{15}=625*((1/5)^{15}-1)/(1/5-1)=625*(1-(1/5)^{15})/(4/5)=

=3125/4*(1-(1/5)^{15})=3125*30517578124/4/30517578125=7629394531/9765625=781.2499999744