Question

The sum of an infinite geometric series is $27$ times the series that results if the first three terms of the original series are removed. What is the value of the series' common ratio

Answer 1

Let $a$ be the first term in the series and $r$ the common ratio. Then the infinite series converges to

$a + ar + ar^2 + ar^3 + ar^4 + ar^5 + \cdots = \dfrac a{1-r}$

Removing the first three terms from the left side effectively multiplies the right side by 27, so

$ar^3 + ar^4 + ar^5 + \cdots = \dfrac{27a}{1-r}$

By elimination,

$a + ar + ar^2 = -\dfrac{26a}{1-r}$

Solve for $r$. We can eliminate $a$ so that

$1 + r + r^2 = -\dfrac{26}{1-r} \\\\ \implies 1-r^3 = -26 \\\\ \implies r^3 - 27 = 0 \\\\ \implies r^3 = 27 \implies \boxed{r=3}$