In a geometric progression, the sum of the first and second terms is 15, and the second and third terms are 30. Find the first 3 terms.
Let us denote by b1 the first term of this geometric progression, and by q the denominator of this geometric progression.
In the condition of the problem it is said that the sum of the first and second members of this progression is 15, and the sum of the second and third members of this progression is 30, therefore, we can compose the following equations:
b1 + b1 * q = 15;
b1 * q + b1 * q ^ 2 = 30.
We solve the resulting system of equations.
Dividing the second equation by the first, we get:
(b1 * q + b1 * q ^ 2) / (b1 + b1 * q) = 30/15;
q = 2.
Substituting the found value q = 2 into the equation b1 + b1 * q = 15, we get:
b1 + b1 * 2 = 15;
3b1 = 15;
b1 = 15/3;
b1 = 5.
b2 = b1 * q = 5 * 2 = 10.
b3 = b2 * q = 10 * 2 = 20.
Answer: The first term is 5, the second term is 10, the third term is 20.