# Find the sum of the first thirty terms of the arithmetic progression given by the formula An = 3n + 2

We find the member of this sequence that comes first.

Substituting the value n = 1 into the formula that sets this sequence, we get:

a1 = 3 * 1 + 2 = 3 + 2 = 5.

We find the member of this sequence, which is in second place.

Substituting the value n = 2 into the formula that sets this sequence, we get:

a2 = 3 * 2 + 2 = 6 + 2 = 8.

Find the difference d of this progression:

d = a2 – a1 = 8 – 5 = 3.

Substituting the sum of the first n members of the arithmetic progression Sn = (2 * a1 + d * (n – 1)) * n / 2 values a1 = 5, d = 3, n = 30 into the formula, we find the sum of the first thirty members of this arithmetic progression:

S30 = (2 * a1 + d * (30 – 1)) * 30/2 = (2 * a1 + d * 29) * 15 = (2 * 5 + 3 * 29) * 15 = (10 + 87) * 15 = 97 * 15 = 1455.

Answer: the sum of the first thirty terms of this arithmetic progression is 1455.