Geometrical progression is given by the following formula:
an+1=an*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,
an=a1*qn-1;
So, for the sixth term in your case, we need to determine q and a1. a1=625; q=125/625, ratio of two successive terms, q=1/5.
a6=625*(1/5)5=625/55=1/5;
Now, for the sum of N terms, from the first to the Nth one.
SN=a1+a2+…+aN=a1+a1*q+a1*q2+…+a1*qN-1=a1*(1+q+…+qN-1)
Let us multiply and divide this expression by (q-1). We will obtain the following:
SN=a1*(q-1)*(1+q+…+qN-1)/(q-1)=a1*(qN-1)/(q-1)
Thus, for the sum of 15 terms we will get:
S15=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