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

sum = a(1-r^n)/(1-r)
http://en.wikipedia.org/wiki/Geometric_series.
solve.

Answer 2

????

Answer 3

in ronncc's formula

a = first term = 21, n = 6 (number of terms), and r = common ratio

Answer 4

is it 1527890?

Answer 5

the terms or of the form a, ar , ar^2, ar^3 .....

6th term = ar^5
r^5 = 6th term / first term

r^5 = 1,240,029 / 21 = 59049
r = 9
so plug these values into the formula:

sum = 21 * 9^6 - 1
------ =
6 - 1

Answer 6

= 2232048

Answer 7

where are you van??