In a decreasing geometric progression, the $$2^{\text{nd}}$$ term is 6. The sum of all infinite terms of the progression is one-eighth of the sum to infinity of the squares of the terms. The sum of the $$1^{\text{st}}$$ and the $$4^{\text{th}}$$ terms is $$\frac{p}{q}$$ where $$p, q$$ are relatively prime to each other. Then the value of $$\left[\frac{p}{q}\right]$$, where $$[x]$$ represents the greatest integer not exceeding $$x$$ is
Correct Answer: 13
Solution
With first term $$A$$ and ratio $$r$$, the condition $$\frac{A}{1-r} = \frac{1}{8} \cdot \frac{A^2}{1-r^2}$$ gives $$A = 8(1 + r)$$, and $$Ar = 6$$ then gives $$4r^2 + 4r - 3 = 0$$, so $$r = \frac{1}{2}$$ and $$A = 12$$. The first and fourth terms add to $$12 + \frac{12}{8} = \frac{27}{2}$$, so $$\left[\frac{p}{q}\right] = \left[13.5\right] = 13$$.
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