Question

What is the sum of a 6–term geometric series if the first term is 7 and the last term is 54,432?

Answer 1

65317

Explanation:

For a geometric series `a+ar+ar^{2}+ar^{3}+cdots+ar^{n-1}` with `n` terms and `r !=1`, the sum is `(a(r^{n}-1))/(r-1)`.

In the given example, `a=7` and `n=6`. We can also set `7*r^{5}=54432` so that `r^{5}=7776` and `r=7776^{1/5}=6`.

Therefore, the sum is `(7*(6^{6}-1))/(6-1)=(7*46655)/5=65317`.