# Simple Arithmetic Series Question

• Oct 10th 2009, 01:17 PM
Viral
Simple 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 PM
e^(i*pi)
Quote:

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

How do I do this?

\$\displaystyle U_n = a+(n-1)d\$

Where

• \$\displaystyle U_n\$ is the nth 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
• Oct 10th 2009, 01:37 PM
Viral
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 PM
e^(i*pi)
Quote:

Originally Posted by Viral
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 >< .

\$\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 PM
Viral
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 PM
e^(i*pi)
Quote:

Originally Posted by Viral
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?

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 PM
Viral
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 AM
e^(i*pi)
Quote:

Originally Posted by Viral
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.

Do you get what I did above with the first three terms? In this case set the numbers you're given here

\$\displaystyle U_1 = a = -8\$

\$\displaystyle U_2 = a+d = -8+d = x^2\$

\$\displaystyle U_3 = a+2d = -8+2d = 17x\$