In an arithmetic series the 20th term is 14 and the 40th term is -6. Find the 10th term.

How do I do this?

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- Oct 10th 2009, 01:17 PMViralSimple Arithmetic Series Question
In an arithmetic series the 20th term is 14 and the 40th term is -6. Find the 10th term.

How do I do this? - Oct 10th 2009, 01:21 PMe^(i*pi)
$\displaystyle U_n = a+(n-1)d$

Where

- $\displaystyle U_n$ is the
*n*th term - $\displaystyle a$ is the first term
- $\displaystyle n$ is the number of terms
- $\displaystyle d$ is the common difference.

$\displaystyle U_{20} = a + (20-1)d = 14$ (eq1)

$\displaystyle U_{40} = a + (40-1)d = -6$ (eq2)

Solve simultaneously. It would be easiest to eliminate a first - $\displaystyle U_n$ is the
- Oct 10th 2009, 01:37 PMViral
Yeah, I've done that, and got $\displaystyle d = 1 \therefore a = -5$.

However, when I do:

$\displaystyle U_{10}=-5+9d=14=v$

$\displaystyle v=-5+9=4$

But the answer is supposed to be 24 >< . - Oct 10th 2009, 01:56 PMe^(i*pi)
$\displaystyle U_{20} > U_{40}$ which means $\displaystyle d<0$

I took equation 2 from equation 1:

$\displaystyle (a+19d) - (a+39d) = 14 - (-6)$

$\displaystyle -20d = 20 \: \therefore \: d = -1$

$\displaystyle a+19(-1) = 14 \: \therefore \: a = 33$

$\displaystyle U_{10} = a + 9d = 33-9 = 24$ - Oct 10th 2009, 02:05 PMViral
I'm now told: The first three terms of an arithmetic series are 5x, 20 and 3x. Find the values of x and hence the values of the three terms.

What are the steps to do this? - Oct 10th 2009, 02:09 PMe^(i*pi)
By definition: $\displaystyle a = 5x$

$\displaystyle U_2 = 20 = a+d $

$\displaystyle U_3 = 3x = a+2d$

You can then solve U_2 and U_3 simultaneously. Eliminate*d*since we don't need to find it

I get $\displaystyle x=5$ and so $\displaystyle a = 25$, $\displaystyle U_2 = 20$ and $\displaystyle U_3 = 15$ - Oct 10th 2009, 02:26 PMViral
Ok, this one is worded really weirdly >< . For this can you just tell me what it's asking, and I'll go from there.

For which values of x would the expression $\displaystyle -8, x^{2} and 17x$ form the first three terms of an arithmetic series. - Oct 11th 2009, 02:48 AMe^(i*pi)