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# Is there another way instead of using trial and error (Quadratic sequences). watch

1. Hi there .

I am working on the below question:

The first 4 terms of a sequence are: 400, 390, 375, 355...
Which term is the first to be negative?

I have worked out the nth term to be -2.5n^2 - 2.5n + 405

I know I need to find the lowest value for n such that

-2.5^2 - 2.5n > 405

I am just using trial and error by subbing in random numbers and working my way through. Is there a quicker/better way to do this?

Thanks
Using quad. formula on -2.5n^2 - 2.5n + 405 = 0 will give n such that it is zero. n + 1 (or if n is not an integer, round up) would be less than zero.
3. Yeah you can bring the 405 back to the other side so you’ve got -2.5n^2 - 2.5n 405>0 and solve it as if it’s any normal quadratic (I.e solve it being equal to zero first). Then draw the graph and mark o where it crosses the x axis. Look at the portion of the graph that is above the x axis, and then you’ve got your values of n for which the sequence will go below zero, then choose the smallest value from this set of values. Hope that kinda makes sense and helps
4. Tn < 0 | Tn = -2.5n^2 - 2.5n + 405.

Now solve. Youll get the answer.
5. (Original post by makin)
Hi there .

I am working on the below question:

The first 4 terms of a sequence are: 400, 390, 375, 355...
Which term is the first to be negative?

I have worked out the nth term to be -2.5n^2 - 2.5n + 405

I know I need to find the lowest value for n such that

-2.5^2 - 2.5n > 405

I am just using trial and error by subbing in random numbers and working my way through. Is there a quicker/better way to do this?

Thanks
Yes, do you know how to solve quadratic inequalities?
6. If you want to the first nth term to be negative then -2.5n^2-2.5n+405<0

you can solve the above inequality to find n.
7. (Original post by makin)
Hi there .

I am working on the below question:

The first 4 terms of a sequence are: 400, 390, 375, 355...
Which term is the first to be negative?

I have worked out the nth term to be -2.5n^2 - 2.5n + 405

I know I need to find the lowest value for n such that

-2.5^2 - 2.5n > 405

I am just using trial and error by subbing in random numbers and working my way through. Is there a quicker/better way to do this?

Thanks
(Original post by Edgemaster)
Using quad. formula on -2.5n^2 - 2.5n + 405 = 0 will give n such that it is zero. n + 1 (or if n is not an integer, round up) would be less than zero.
(Original post by El_Swego)
Yeah you can bring the 405 back to the other side so you’ve got -2.5n^2 - 2.5n 405>0 and solve it as if it’s any normal quadratic (I.e solve it being equal to zero first). Then draw the graph and mark o where it crosses the x axis. Look at the portion of the graph that is above the x axis, and then you’ve got your values of n for which the sequence will go below zero, then choose the smallest value from this set of values. Hope that kinda makes sense and helps
(Original post by UziGun)
Tn < 0 | Tn = -2.5n^2 - 2.5n + 405.

Now solve. Youll get the answer.
(Original post by Muttley79)
Yes, do you know how to solve quadratic inequalities?
(Original post by Mystery.)
If you want to the first nth term to be negative then -2.5n^2-2.5n+405<0

you can solve the above inequality to find n.
Thanks for all the help. I've worked it down and that narrowed it but still used trial and error to get to 13. It was a lot quicker than what I had though.

8. 5n^2+5n-810>0

Just solve this using the root of quadratic equation formula.

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Updated: February 4, 2018
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