Determine the sum of the first 20 terms of the arithmetic progression a, 2a, 3a, 4a, 5a.

To find the sum of the first 20 terms of the arithmetic progression a, 2a, 3a, 4a, 5a, we first need to find the common difference between the terms.

The common difference between consecutive terms in an arithmetic progression is the difference between any two consecutive terms. In this case, the common difference is 2a - a = a.

The formula to find the sum of the first n terms of an arithmetic progression is given by:
S_n = n/2 * (2a + (n-1)d)

where:
S_n = sum of the first n terms
n = number of terms
a = first term
d = common difference

In this case, n = 20, a = a, and d = a. Plugging these values into the formula, we get:
S_20 = 20/2 * (2a + (20-1)a)
S_20 = 10 * (2a + 19a)
S_20 = 10 * 21a
S_20 = 210a

Therefore, the sum of the first 20 terms of the arithmetic progression a, 2a, 3a, 4a, 5a is 210a.

The sum of the first n terms of an arithmetic progression is given by the formula:

$$S_n = \frac{n}{2}(2a + (n-1)d)$$

where a is the first term, d is the common difference, and n is the number of terms.

In this case, a = a, d = a, and n = 20. Substituting these values into the formula, we get:

$$S_{20} = \frac{20}{2}(2a + (20-1)a)$$

$$S_{20} = 10(2a + 19a)$$

$$S_{20} = 10(21a)$$

$$S_{20} = 210a$$

Therefore, the sum of the first 20 terms of the arithmetic progression a, 2a, 3a, 4a, 5a is 210a.