# Arithmetic series - stuck!

• May 22nd 2012, 11:42 AM
Arithmetic series - stuck!
Hi all, just joined and introduced myself in the lobby.

I have two questions which I have not been able to do the workings out for. I understand the formula to calculate arithmetic series, but I am stumped on these - hope you can help.

1. The sum of an arithmetic series is 270. The common difference is 1 and the first term is 4. Calculate the number of terms in the series.

2. The first term of an arithmetic series is 16. The 30th term is 100. Calculate S30.

Can anybody show me the process they would go through?

Thanks

• May 22nd 2012, 12:11 PM
skeeter
Re: Arithmetic series - stuck!
sum of an arithmetic series ...

$\displaystyle S_n = a_1 + (n-1)d$

you know $\displaystyle S_n$ , $\displaystyle d$, and $\displaystyle a_1$ ... so, what's the problem with finding $\displaystyle n$ ?
• May 22nd 2012, 05:46 PM
Soroban
Re: Arithmetic series - stuck!

You say you understand the formulas.
I don't understand your difficulty.

Quote:

1. The sum of an arithmetic series is 270.
The common difference is 1 and the first term is 4.
Calculate the number of terms in the series.

The sum is the first n terms of an arithmetic series: .$\displaystyle S_n \:=\:\frac{n}{2}\big[2a_1 + (n-1)d\big]$

We are given: .$\displaystyle S_n = 270,\;a_1=4,\;d = 1$

We have: .$\displaystyle \frac{n}{2}\big[2(4) + (n-1)1\big] \:=\:270$

$\displaystyle \text{Solve for (positive) }n.$

Quote:

2. The first term of an arithmetic series is 16.
The 30th term is 100.
Calculate S
30

$\displaystyle \text{The }n{th}\text{ term is: }\:a_n \:=\:a_1 + (n-1)d$

We are given: .$\displaystyle a_1 = 16,\;n=30,\;a_{30} = 100$

We have: .$\displaystyle 16 + 29d \:=\:100 \;\hdots\;\text{Solve for }d.$

$\displaystyle \text{Then determine }S_{30}$
• May 23rd 2012, 03:29 AM
skeeter
Re: Arithmetic series - stuck!
Quote:

Originally Posted by skeeter
sum of an arithmetic series ...

$\displaystyle S_n = a_1 + (n-1)d$ ... nope, thinking sum, putting down nth term. no more drinking and deriving.

mea culpa.
• May 27th 2012, 10:12 PM
Re: Arithmetic series - stuck!
Thanks for your responses:

Don't know how you write your formula's in that font, but I understand it up to this point:

1.
n/2(8+(n-1)1),

n/2(8+(n-1)) = 270

Then what is the next step?

Thanks for your patience!

PD
• May 27th 2012, 11:13 PM
linalg123
Re: Arithmetic series - stuck!
expand out the brackets and use the quadratic formula.
$\displaystyle \frac{n}{2}(8+(n-1)) = 270$
$\displaystyle n(8+(n-1)) = 540$
$\displaystyle 8n+ n(n-1) = 540$
$\displaystyle n^2 + 7n - 540 = 0$
now use the quadratic formula

remember the quadratic formula will give you two solutions, only the positive one is a feasible answer.
• May 28th 2012, 12:01 PM
Re: Arithmetic series - stuck!
thanks Iinalg123

understand up to 7n+n2=540

couldn't you then divide by 7 on both sides?

With n2 + 7n - 540 is there an easy way to work the quadratic formula out?

Can someone please tell me how to write the formula's in the fonts above - thanks.

PD
• May 28th 2012, 11:11 PM
linalg123
Re: Arithmetic series - stuck!
Lets say you have a quadratic $\displaystyle ax^2 + bx + c = 0$

the quadratic formula says the solutions are $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

If you use this formula for your quadratic you will get your solution for x.
• May 29th 2012, 11:37 AM
Re: Arithmetic series - stuck!
ok, so I have the following:

-7 + SQRT 7 squared - 4(1)(540)
divided by
2(1)

Then:
-7 + SQRT 49 -2160
divided by
2

Then:
-7 + SQRT - 2111
divided by
2

Then:
-7 + 45.95
divided by
2

Then:
38.95
divided by
2

Then:
19.475, which rounded up is 20, which is the text book answer - thank you.
• May 29th 2012, 06:58 PM
linalg123
Re: Arithmetic series - stuck!
Close, however you didn't convert it to the form $\displaystyle ax^2+bx+c=0$, you used $\displaystyle ax^2 + bx = c$
So, if we convert it first we have $\displaystyle n^2 + 7n -540 = 0$ (note the negative sign for c)

So, $\displaystyle n = \frac{-7 \pm \sqrt{7^2 - 4(1)(-540)}}{2(1)}$

i.e $\displaystyle n = \frac{-7 \pm \sqrt{49 +2160)}}{2}$ (note the sign chance as negative * negative is positive.)

$\displaystyle n = \frac{-7 \pm \sqrt{2209}}{2}$

$\displaystyle n = \frac{-7 \pm 47}{2}$

now as we only considering the positive answer $\displaystyle n = \frac{40}{2} = 20$
• May 29th 2012, 09:59 PM
Re: Arithmetic series - stuck!
With regard to the second question:

Quote:

The first term of an arithmetic series is 16. The 30th term is 100. Calculate S30
I have the following formula:

$\displaystyle S_n \:=\:\frac{n}{2}\big[2a + (n-1)d\big]$

so

$\displaystyle S_1 \:=\:\frac{1}{2}\big[2(16) + (1-1)d\big]$ $\displaystyle \:=\:\frac{1}{2}\big[32 + d\big]$ $\displaystyle \:=\:\big[16 + \frac{1}{2}d\big]$

and

$\displaystyle S_3_0 \:=\:\frac{30}{2}\big[2(100) + (30-1)d\big]$ $\displaystyle \:=\:\frac{30}{2}\big[200 + 29d\big]$ $\displaystyle \:=\:\15\big[200 + 29d\big]$ $\displaystyle \:=\:\big[3000 + 435d\big]$

Where from here? Can't help feeling I am going about this in the wrong way.

PD
• May 29th 2012, 10:02 PM