# How do you find the sum of the infinite geometric series a1 = 42 and r = 6/5?

It diverges and has no real sum since $| r | > 1$
Since $\setminus | r | = \frac{6}{5} > 1$, it implies that the series does not converge and hence it diverges to infinity and has no real sum.
That is ${\sum}_{n = 1}^{\infty} 42 \cdot {\left(\frac{6}{5}\right)}^{n - 1} = \infty \notin \mathbb{R}$