In a geometric progression, the sum of the first and second terms is 16, and the sum of the second and third terms is 48. Find the first three terms of this progression.
Given: (bn) – geometric progression;
b1 + b2 = 16, b2 + b3 = 48;
Find: b1, b2, b3 -?
Formula of the nth term of a geometric progression:
bn = b1 * q ^ (n-1),
where b1 is the first term of the progression, q is the denominator of the progression.
Let us express the second and third terms of the progression through the formula for the nth term:
b2 = b1 * q ^ (2-1) = b1 * q;
b3 = b1 * q ^ (3-1) = b1 * q ^ 2.
Then b1 + b2 = b1 + b1 * q = b1 (1 + q);
b2 + b3 = b1 * q + b1 * q ^ 2 = b1 * q (1 + q).
Next, we solve the system of equations:
b1 (1 + q) = 16, (1)
b1 * q (1 + q) = 48 (2)
Let us express from (1) the equation b1:
b1 = 16 / (1 + q).
Substitute the resulting expression into (2) the equation:
16q * (1 + q) / (1 + q) = 48;
Reducing the same factors in the numerator and denominator and get:
16q = 48;
q = 3.
Now we find the required members of the given progression:
b1 = 16 / (1 + q) = 16 / (1 + 3) = 4;
b2 = b1 * q = 4 * 3 = 12;
b3 = b1 * q ^ 2 = 4 * 3 ^ 2 = 36.
Answer: b1 = 4, b2 = 12, b3 = 36.