Natural numbers are members of a geometric progression with a natural denominator.

Natural numbers are members of a geometric progression with a natural denominator. The sum of the first three terms of this progression is 31. Find the fifth term of the progression.

Let us denote by b1 the first term of this geometric progression, and by q the denominator of this geometric progression.

Then the second and third terms of this progression will be equal:

b2 = b1 * q;

b3 = b1 * q ^ 2.

According to the condition of the problem, the sum of the first three members of this progression is 3, therefore, we can write the following ratio:

b1 + b1 * q + b1 * q ^ 2 = 31;

b1 * (1 + q + q ^ 2) = 31.

Since the number 31 is prime, and the numbers b1 and 1 + q + q ^ 2 are naturals, the following equalities must hold:

b1 = 1;

1 + q + q ^ 2 = 31.

From the second relation, by Vieta’s theorem, we find q:

q ^ 2 + q – 30 = 0;

q1 = 5;

q2 = -6.

Since q is a natural number, the value q = -6 is not suitable.

Find b5:

b5 = b1 * q ^ 4 = 1 * 5 ^ 4 = 625.

Answer: the fifth term of the progression is 625.