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?

1,395,030

1,461,460

1,527,890

1,594,320

Answer 1

First, you have to get a handle on an expression for that geometric series.

Take that last (6th) term and divide it by the first to see what number is "something to the 5th power". That is because your first term is 21 times x^0 and you have to find "x". Your last term is 21 times x^5.

1240029 / 21 = 59049 and 59049 is 9^5 so we can get our geometric expression now.

S1 = 21(9^0 + 9^1 + ... + 9^4 + 9^5)

S2 = 9^0 + 9^1 + ... + 9^4 + 9^5 = (9^6 - 1)/(9 - 1)

So, S1 = (21)(9^6 - 1)/(9 - 1)

This is now quite easy to work with, so once you get 9^6 (use a calculator!) you will have no problem!

Answer 2

thx for the medal, but take some time to go through this to see if you get it. I'll hang around.