# Arithmetic Series Question

• Nov 18th 2012, 12:05 AM
Vince604
Arithmetic Series Question
The sum of the first 5 terms of an arithmetic series is 85, and the sum of the first 6 terms is 123. Write the first 3 terms of the series.

Okay i tried, but ended up with strange numbers.

formulas:
tn=t1+(n-1)d
sn=n/2(2(t1)+(n-1)d)

• Nov 18th 2012, 12:16 AM
MarkFL
Re: Arithmetic Series Question
Using the summation formula, we find:

$\displaystyle S_5=\frac{5}{2}(2t_1+4d)=85$

$\displaystyle S_5=3(2t_1+5d)=123$

The two equations may be written:

$\displaystyle t_1+2d=17$

$\displaystyle 2t_1+5d=41$

Now, solve this system to determine $\displaystyle t_1$ and $\displaystyle d$.
• Nov 18th 2012, 12:46 AM
Vince604
Re: Arithmetic Series Question
thanks Mark, helpful as always. i try later, too tired now...
• Nov 18th 2012, 03:32 PM
Vince604
Re: Arithmetic Series Question
Quote:

Originally Posted by MarkFL2
Using the summation formula, we find:

$\displaystyle S_5=\frac{5}{2}(2t_1+4d)=85$

$\displaystyle S_5=3(2t_1+5d)=123$

The two equations may be written:

$\displaystyle t_1+2d=17$

$\displaystyle 2t_1+5d=41$

Now, solve this system to determine $\displaystyle t_1$ and $\displaystyle d$.

I actually wrote it out exactly like you did, but then i got help from another site as well and the person said that i could bring the fraction (5/2 & 6/2) over and use it to divide with 85 and 123. whew. I made it seem so difficult lol.

I blame my teacher for not showing the proper steps!!!

Oh and the two equations with 17 and 41 confuses me, so i didn't know what to do with them.
• Nov 18th 2012, 03:44 PM
MarkFL
Re: Arithmetic Series Question
Given the system:

(1) $\displaystyle t_1+2d=17$

(2) $\displaystyle 2t_1+5d=41$

I would multiply (1) by -2, then add the two equations to eliminate $\displaystyle t_1$:

$\displaystyle d=7$

and then using (1), we find:

$\displaystyle t_1=3$
• Nov 18th 2012, 03:48 PM
Vince604
Re: Arithmetic Series Question
Quote:

Originally Posted by MarkFL2
Given the system:

(1) $\displaystyle t_1+2d=17$

(2) $\displaystyle 2t_1+5d=41$

I would multiply (1) by -2, then add the two equations to eliminate $\displaystyle t_1$:

$\displaystyle d=7$

and then using (1), we find:

$\displaystyle t_1=3$

OHHH, so you try to eliminate what you can to solve right?
• Nov 18th 2012, 03:55 PM
MarkFL
Re: Arithmetic Series Question
Yes, elimination is one way, substitution is another. I tend to favor elimination for small systems like this.
• Nov 18th 2012, 04:04 PM
Vince604
Re: Arithmetic Series Question
Thanks