Arithmetic Series help (AS Level)

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CathyLou
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I'm totally stuck on the following question and so I'd very very grateful if someone could please tell me how to work it out.

The first three terms of an arithmetic series are (12-P), 2P and (4P-5) respectively, where P is a constant.

(a) Find the value of P.

(b) Show that the sixth term of the series is 50.

(c) Find the sum of the first 15 terms of the series.

(d) Find how many terms of the series have a value of less than 400.


Thank you.

Cathy
 
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Present us some of your work. Write down the expression for the general term of an arithmetic sequence, and everything should be more clear. Set up a few equations, and see where they'll bring you.

Edit: late again. :smile:
 
Before I posted I had written the following in response to part a:

an = a1 + (n - 1)d

d = 2p - (12 - p)

d= (4p - 5) - 2p

2p - (12 - p) = (4p - 5) - 2p

4p = (4p - 5) + (12 - p)

Then I got d = 7 but I got confused over the value of p.

Am I working on the right lines?

Thanks for replying by the way! :smile:

Cathy
 
radou said:
No, p should equal 7.

Oh, okay, I get where I went wrong. Thank you.

So, for part b do I just substitute in p to the equation to find d?

Cathy
 
CathyLou said:
Oh, okay, I get where I went wrong. Thank you.

So, for part b do I just substitute in p to the equation to find d?

Cathy

Yes, you do. I.e., d must equal a3 - a2, and a2 - a1, doesn't matter which difference you take.
 
Thanks for your help.

I've now worked out parts a, b and c but I'm still unsure how to do d. Any hints would be really appreciated.

Cathy
 
Basically, you can solve part (d) unformally by trying to plug in different numbers n into an = a1 + (n - 1)*d and see which one is the greatest term which is less than 400.
 
radou said:
Basically, you can solve part (d) unformally by trying to plug in different numbers n into an = a1 + (n - 1)*d and see which one is the greatest term which is less than 400.

Okay, thanks very much! :biggrin:

Cathy