- #1

- 18

- 0

find a and r

[tex]a+ar+ar^2=7[/tex]

[tex]a^3+a^3r^3+a^3r^6=73[/tex]

[tex]a+ar+ar^2=7[/tex]

[tex]a^3+a^3r^3+a^3r^6=73[/tex]

- Thread starter santa
- Start date

- #1

- 18

- 0

find a and r

[tex]a+ar+ar^2=7[/tex]

[tex]a^3+a^3r^3+a^3r^6=73[/tex]

[tex]a+ar+ar^2=7[/tex]

[tex]a^3+a^3r^3+a^3r^6=73[/tex]

- #2

- 488

- 2

http://www.imf.au.dk/kurser/algebra/E05/GBintro.pdf [Broken]find a and r

[tex]a+ar+ar^2=7[/tex]

[tex]a^3+a^3r^3+a^3r^6=73[/tex]

Last edited by a moderator:

- #3

- 18

- 0

OK but the solution where

- #4

- 695

- 2

1 + 2 + 4 = 7,

1 + 8 + 64 = 73.

- #5

- 441

- 0

Not sure if there is more than one solution, but I found that a = 1 and r =2 works.

- #6

- 836

- 13

Then you can solve for 'a' or 'r', and substitute into the other equation. It's not pretty, but it will work!

I remember seeing a very similar problem in high school, must see if I can find it again.

In fact, you could try by summing geometric equations.

a + ar + ar^2 = a(1 - r^3)/(1 - r) = 7

a^3 + a^3.r^3 + a^3.r^6 = a^3(1 - r^9)/(1 - r^3) = 73

Divide the one sum by the other and see where it leads you.

- Last Post

- Replies
- 5

- Views
- 3K

- Replies
- 8

- Views
- 2K

- Replies
- 2

- Views
- 4K

- Replies
- 13

- Views
- 807

- Replies
- 3

- Views
- 2K

- Replies
- 4

- Views
- 138

- Replies
- 1

- Views
- 2K

- Replies
- 3

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 5

- Views
- 9K