Question

What is the sum of a 6-term geometric series if the first term is 21 and the last term is 1,240,029?

Answer 1

you need to find the common ratio r

1'll assume the 6th term is 1240029

a term in a geometric series if found using

\[T_{n} = ar^{n -1} \]

in your question n = 6 and a = 21

substitute and then solve for r

once you have r the sum of a geometric series can be found using

\[S_{n} = \frac{a(r^n - 1)}{r -1}\]

again a = 21 and n = 6 and you need r from above