Let [tex]a[/tex] be the first term in the series and [tex]r[/tex] the common ratio. Then the infinite series converges to
[tex]a + ar + ar^2 + ar^3 + ar^4 + ar^5 + \cdots = \dfrac a{1-r}[/tex]
Removing the first three terms from the left side effectively multiplies the right side by 27, so
[tex]ar^3 + ar^4 + ar^5 + \cdots = \dfrac{27a}{1-r}[/tex]
By elimination,
[tex]a + ar + ar^2 = -\dfrac{26a}{1-r}[/tex]
Solve for [tex]r[/tex]. We can eliminate [tex]a[/tex] so that
[tex]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}[/tex]
