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 = \frac 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 = (27a)/(1-r)`

By elimination,

`a + ar + ar^2 = -(26a)/(1-r)`

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

`1 + r + r^2 = -(26)/(1-r) \\\\ \implies 1-r^3 = -26 \\\\ \implies r^3 - 27 = 0 \\\\ \implies r^3 = 27 \implies \boxed{r=3}`