# Math Help - Arithmetic Series Question

1. ## 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)

2. ## Re: Arithmetic Series Question

Using the summation formula, we find:

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

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

The two equations may be written:

$t_1+2d=17$

$2t_1+5d=41$

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

3. ## Re: Arithmetic Series Question

thanks Mark, helpful as always. i try later, too tired now...

4. ## Re: Arithmetic Series Question

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

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

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

The two equations may be written:

$t_1+2d=17$

$2t_1+5d=41$

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

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.

5. ## Re: Arithmetic Series Question

Given the system:

(1) $t_1+2d=17$

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

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

$d=7$

and then using (1), we find:

$t_1=3$

6. ## Re: Arithmetic Series Question

Originally Posted by MarkFL2
Given the system:

(1) $t_1+2d=17$

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

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

$d=7$

and then using (1), we find:

$t_1=3$
OHHH, so you try to eliminate what you can to solve right?

7. ## Re: Arithmetic Series Question

Yes, elimination is one way, substitution is another. I tend to favor elimination for small systems like this.

Thanks