**Solution:**

Let's consider an infinite geometric progression with the first term $a$ and the common ratio $r$. The sum of an infinite geometric progression can be calculated using the formula:

$$S = \frac{a}{1-r}$$

Given that the sum of the infinite geometric progression is 10, we can write the equation as:

$$10 = \frac{a}{1-r}$$

To find the possible integral values of the first term, we need to consider the values of $a$ that satisfy this equation.

**Case 1: $1-r \neq 0$**

If $1-r \neq 0$, we can rearrange the equation as:

$$a = 10(1-r)$$

The first term $a$ can take any integral value as long as $1-r \neq 0$. This means $r$ cannot be equal to 1.

**Case 2: $1-r = 0$**

If $1-r = 0$, the common ratio $r$ is equal to 1. In this case, the geometric progression becomes a constant sequence with all terms equal to the first term $a$. Since the sum of the sequence is 10, we have:

$$10 = a + a + a + \ldots = a + a(1) + a(1)^2 + \ldots$$

Simplifying this equation, we get:

$$10 = a(1 + 1 + 1 + \ldots) = a\cdot\infty$$

Since the sum of an infinite number of terms cannot be a finite number, this equation has no valid solution.

Therefore, we only need to consider values of $a$ in Case 1, where $r \neq 1$.

**Finding the integral values of $a$**

To find the integral values of $a$, we can express $10(1-r)$ as a product of two integers, where one integer is a factor of 10 and the other is a factor of $1-r$.

The factors of 10 are: 1, 2, 5, and 10.

For each factor of 10, we need to find the corresponding factor of $1-r$ that gives an integral value of $a$. This can be done by rearranging the equation $a = 10(1-r)$ as:

$$1-r = \frac{a}{10}$$

We can substitute each factor of 10 and solve for $r$. If $r$ is an integer, then $a$ will also be an integer.

- For $r = -9$, we have $1-r = 10$. This gives $a = 1$.
- For $r = -8$, we have $1-r = 9$. This gives $a = 2$.
- For $r = -5$, we have $1-r = 6$. This gives $a = 5$.
- For $r = -4$, we have $1-r = 5$. This gives $a = 10$.

Therefore, the possible integral values of $a$ are 1, 2, 5, and 10.

**Sum of all possible integral values of $a$**

To find the sum of all possible integral values of $a$, we simply add them up