Let $$a_n be n^th the 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

The correct answer is 54.

Let the terms of the decreasing infinite geometric progression be represented as:

$a_1=a$

$a_2=ar$

$a_3=a{r}^2$

where $a$ is the first term and $r$ is the common ratio. Since it is a decreasing infinite geometric progression, we know that $\left|r\right|<1$.

We are given the following two equations:

1) $a_1+a_2+a_3=52$

2) $a_1{a}_2+a_2{a}_3+a_3{a}_1=624$

Substituting the terms into the first equation, we get:

$a+ar+a{r}^2=52$

$a(1+r+r^2)=52$

Substituting the terms into the second equation, we get:

$\left(a\right)\left(ar\right)+\left(ar\right)\left(a{r}^2\right)+\left(a{r}^2\right)\left(a\right)=624$

$a^{2}r+a^2{r}^3+a^2{r}^2=624$

$a^{2}r(1+r^2+r)=624$

Now, divide the second simplified equation by the first simplified equation:

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

$ar=12$

From this, we can express $a$ in terms of $r$:

$a=\frac{12}{r}$

Substitute $ar=12$ back into the first equation:

$a+12+a{r}^2=52$

$a+a{r}^2=40$

Substitute $a=\frac{12}{r}$ into this equation:

$\frac{12}{r}+\left(\frac{12}{r}\right)r^2=40$

$\frac{12}{r}+12r=40$

Divide the entire equation by 4:

$\frac{3}{r}+3r=10$

Multiply by $r$ and rearrange to form a quadratic equation:

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

Factorizing the quadratic equation:

$3r^2-9r-r+3=0$

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

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

So, the possible values for $r$ are:

$r=\frac{1}{3}$ or $r=3$

Since the geometric progression is decreasing, the common ratio must satisfy $\left|r\right|<1$. Therefore, we take:

$r=\frac{1}{3}$

Now, substitute $r$ back to find $a$:

$a=\frac{12}{(1/3)}=36$

The sum of an infinite geometric progression is given by the formula:

$S_{\infty }=\frac{a}{1-r}$

Substitute the values of $a$ and $r$:

$S_{\infty }=\frac{36}{1-\frac{1}{3}}$

$S_{\infty }=\frac{36}{\left(\frac{2}{3}\right)}$

$S_{\infty }=36\times \frac{3}{2}=54$

Therefore, the sum of this geometric progression is 54.