Question

If \( 7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \cdots \infty \text{ terms}, \) then \( \alpha \) is equal to

Hint:

For solving infinite geometric series, use the formula for the sum: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio.

Answer 1

We are given an infinite series. The general form of the given series is: \[ 7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \cdots \] This series is in the form of a geometric series with the first term \( 5 \) and common ratio \( \frac{1}{7} \), with \( \alpha \) being the factor that is progressively increasing. To solve for \( \alpha \), we equate the sum of the series to 7: \[ 7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \cdots \] Simplifying this and solving for \( \alpha \), we find that \( \alpha = \frac{1}{7} \).