# arithmetic sequence problem

Printable View

• Oct 25th 2008, 01:14 AM
listeningintently
arithmetic sequence problem
Hello :D

Here's the question I need a hand with:

You have a 10-year plan to raise money for university. You have the choice of two jobs:

The first has a commencing salary of \$30 000 per year with an annual increment of \$1000

The second offers a commencing salary of \$31 000 per year with a six monthly increment of \$500

Over 10 years which job would pay the most and by how much?

Thanks in advance, all help is much appreciated.
• Oct 25th 2008, 04:53 AM
HallsofIvy
Quote:

Originally Posted by listeningintently
Hello :D

Here's the question I need a hand with:

You have a 10-year plan to raise money for university. You have the choice of two jobs:

The first has a commencing salary of \$30 000 per year with an annual increment of \$1000

The second offers a commencing salary of \$31 000 per year with a six monthly increment of \$500

Over 10 years which job would pay the most and by how much?

Thanks in advance, all help is much appreciated.

YOu say this is an arithmetic sequence. Don't you know any formulas for summing arithmetic sequences?
• Oct 25th 2008, 06:09 PM
listeningintently
haha...funny...

Yes, I am aware of these formulas, I simply wanted a hand with how to figure out the second case as the answers in my book differed to the one I received.
• Oct 25th 2008, 07:53 PM
ThePerfectHacker
Quote:

Originally Posted by listeningintently
The first has a commencing salary of \$30 000 per year with an annual increment of \$1000

At the 0-th zero you have $30000+(0)(1000)$
At the 1-st zero you have $30000+(1)(1000)$
At the 2-nd zero you have $30000+(2)(1000)$
...
In general at the 10th year you have $30000+(10)(1000)$

Quote:

The second offers a commencing salary of \$31 000 per year with a six monthly increment of \$500
Similar reasoning says that at the 10-th you have $31000 + (10)(500)$

Now which one is bigger?