Let $$a_{n}$$ be the $$n^{th}$$term of a decreasing infinite geometric progression. If $$a_{1}+a_{2}+a_{3}=52$$ and $$a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=624$$, then the sum of this geometric progression is

Let the first term be a and the common ratio be r.

Given

$$a+a r + a r^2 = 52 \quad\text{and}\quad a^2 r + a^2 r^2 + a^2 r^3 = 624.$$

From the first equation $$a(1+r+r^2)=52$$, so $$a=\dfrac{52}{1+r+r^2}$$. 

Substitute into the second:

$$\frac{52^2}{(1+r+r^2)^2}*(r+r^2+r^3)=624$$

$$\frac{r}{(1+r+r^2)}=\frac{3}{13}$$

$$3r^2-10r+3=0$$

$$(3r-1)(r-3)=0$$

Simplify to get an equation in (r); its real solutions are $$r=\tfrac{1}{3}$$ and $$r=3$$. For an infinite geometric progression, we must have (|r|<1), so $$r=\dfrac{1}{3}$$.

Now
$$a=\frac{52}{1+\tfrac{1}{3}+\tfrac{1}{9}}=\frac{52}{13/9}=36$$

The sum of the infinite progression is

$$ S=\frac{a}{1-r}=\frac{36}{1-\tfrac{1}{3}}=\frac{36}{2/3}=54.$$