Arithmetic series and Gauss method

[Simonsky]

came up with this example and can't seem to work out how the formula is simplified into a quadratic

X starts a job on £9000 per year which increases by an increment of £1000 a year.

The number of years that have elapsed when her total earnings are £100,000 is given by

S =1/2n[ 2a + (n-1)d] where S=100,000, a = 9000 and d = 1000 This gives: 1/2n[2 x 9000 + 1000(n-1)]

This simplifies to: n^2 + 17n - 200 = 0

Help explaining that simplification appreciated!

 

[Deleted member 4993]

came up with this example and can't seem to work out how the formula is simplified into a quadratic

X starts a job on £9000 per year which increases by an increment of £1000 a year.

The number of years that have elapsed when her total earnings are £100,000 is given by

S =1/2n[ 2a + (n-1)d] where S=100,000, a = 9000 and d = 1000 This gives: 1/2n[2 x 9000 + 1000(n-1)]

This simplifies to: n^2 + 17n - 200 = 0

Help explaining that simplification appreciated!

How did you get that?

How do you know that is correct?

 

[Dr.Peterson]

This gives: 1/2n[2 x 9000 + 1000(n-1)]

This simplifies to: n^2 + 17n - 200 = 0

Help explaining that simplification appreciated!

I think you are saying that someone told you that the equation (1/2)n[2(9000) + 1000(n-1)] = 100,000 simplifies to n^2 + 17n - 200 = 0, and you are asking how to do that simplification.

First use the distributive property to expand the left side; then divide all terms by the common factor, 500.

If you have trouble doing that, please show your work so we can see if you are going wrong somewhere.

 

[Simonsky]

Dr. Peterson -many thanks, it was the division by the common factor that I was missing!

Thanks for clarifying that for me!