Question

The sum of the first three terms of an infinite geometric progression, with common ratio less than one, is 56. If 1, 7 and 21 are subtracted from its first, second and third term, respectively, then these three terms are in the arithmetic progression. The common ratio of the progression is:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{2}{3} \)

Hint:

When given a geometric progression with conditions that the terms form an arithmetic progression, use the property of arithmetic progressions to relate the terms and solve for the common ratio.

Answer 1

The Correct Option is A

Let the first term of the geometric progression be \(a\) and the common ratio be \(r\). The sum of the first three terms is given by: \[ S_3 = a + ar + ar^2 = 56 \] From the condition of arithmetic progression, we know: \[ a - 1, \quad ar - 7, \quad ar^2 - 21 \quad \text{are in arithmetic progression} \] For three terms to be in arithmetic progression, the middle term must be the average of the first and third terms: \[ 2(ar - 7) = (a - 1) + (ar^2 - 21) \] Expanding and simplifying the equation, we solve for \(r\) and \(a\). After solving, we find that the common ratio \(r\) is \( \frac{1}{2} \).