The sum of the first 8 terms of a GP is five times the sum of its first 4 terms. If r ≠ 1 is the common ratio, then what is the number of possible real values of r?

The number of possible real values for the common ratio, r, is 3.

Detailed Explanation

Let's denote the sum of the first n terms of a geometric progression (GP) as S_n. The formula for the sum of a GP is:

\(S_n = \frac{a(r^n - 1)}{r - 1}\)

where a is the first term and r is the common ratio.

According to the problem, the sum of the first 8 terms is five times the sum of the first 4 terms. This can be written as:

\(S_8 = 5S_4\)

Substituting the formula for S_n into this equation gives us:

\(\frac{a(r^8 - 1)}{r - 1} = 5 \cdot \frac{a(r^4 - 1)}{r - 1}\)

Since the common ratio r ≠ 1, the term (r - 1) is non-zero. Also, we can assume the first term a ≠ 0, because if a = 0, the entire series is 0, and the condition holds for any r, which doesn't make sense in this context. Therefore, we can cancel out a and (r - 1) from both sides of the equation, leaving:

\(r^8 - 1 = 5(r^4 - 1)\)

We can factor the term (r⁸ - 1) using the difference of squares formula, (x² - y²) = (x - y)(x + y), where x = r⁴ and y = 1:

\((r^4 - 1)(r^4 + 1) = 5(r^4 - 1)\)

Now, we can solve this equation by considering two possible cases:

Case 1: \(r^4 - 1 = 0\)

In this case, we have \(r^4 = 1\). The real solutions for r are r = 1 and r = -1. Since the problem states that r ≠ 1, the only valid solution from this case is r = -1.

Case 2: \(r^4 - 1 \neq 0\)

If \(r^4 - 1\) is not zero, we can divide both sides of the equation by \((r^4 - 1)\):

\(r^4 + 1 = 5\)

\(r^4 = 4\)

To find the real values of r, we take the fourth root of 4:

\(r = \pm \sqrt[4]{4} = \pm \sqrt{\sqrt{4}} = \pm \sqrt{2}\)

So, the two solutions from this case are r = √2 and r = -√2.

Conclusion

By combining the solutions from both cases, the possible real values for r are -1, √2, and -√2. Therefore, there are a total of 3 possible real values for the common ratio r.