In a geometric progression (Bn), the denominator of which is positive, b1 * b2 = 1/27, a b3 * b4 = 3

In a geometric progression (Bn), the denominator of which is positive, b1 * b2 = 1/27, a b3 * b4 = 3. Find the sum of the first four terms of this progression.

1. A geometric progression is given.

b1, b2, b3, b4 are its first, second, third and fourth terms.

Let k be the denominator of the progression.

Then b3 = b2 * k = b1 * k * k.

b4 = b3 * k = b2 * k * k.

2. It is known that b1 * b2 = 1/27.

Find k.

b3 * b4 = b1 * k * k * b2 * k * k.

1/27 * k * k * k * k = 3.

k * k * k * k = 81.

Since, by hypothesis, k> 0, it means that k = 3.

3.b2 = b1 * k = 3 * b1.

So b1 * b2 = b1 * b1 * 3 = 1/27.

b1 * b1 = 1/81.

Then b1 = 1/9, b2 = 3 * 1/9 = 1/3, b3 = 1/3 * 3 = 1, b4 = 1 * 3 = 3.

4. Find the amount.

1/9 + 1/3 + 1 + 3 = 4 4/9.

Answer: The sum is 4 4/9.