The sum of the first and second terms of the geometric progression is 35 more than the sum of the second

The sum of the first and second terms of the geometric progression is 35 more than the sum of the second and third terms, equal to 105. Find the first term and the denominator of this progression.

You are given a progression b1, b2. в3 with the denominator g.

b1 + b2 = b1 + b1 * g = b1 * (1 + g) = 105 + 35 = 140. (1)

b2 + b3 = b1 * g + b1 * g ^ 2 = b1 * g * (1 + g) = 105.

b1 * (1 + g) = 140; в1 * g * (1 + g) = 105. Divide the second equality by the first, we get:

в1 * g * (1 + g) / в1 * (1 + g) = g = 105/140 = 0.75.

Let us substitute in equality (1) the obtained value of the denominator g = 0.75. (g <1, decreasing progression).

b1 * (1 + g) = b1 * (1 + 0.75) = b1 * 1.75 = 140.

From where we define the first term of the progression in1:

b1 = 140 / 1.75 = 80.