The Purplemath Forums

by **maroonblazer** on Sun Nov 13, 2011 4:57 pm

Hi,
I'm trying to figure out how they simplified this:
$[a_1 + d]+[a_1+(n-2)d]$

to this:
$2a_1 + (n-1)d$

I get how the two $a_1$ s get to $2a_1$ but I can't figure out how they're getting the two $d$ s two just one.

Help?

Thanks in advance,
mb

**Sponsor**

by **stapel_eliz** on Sun Nov 13, 2011 11:20 pm

maroonblazer wrote:I'm trying to figure out how they simplified this:
$[a_1 + d]+[a_1+(n-2)d]$

to this:
$2a_1 + (n-1)d$

What did you get when you multiplied out the second bit? You got to here:

. . . . . $a_1\, +\, d\, +\, a_1\, +\, nd\, -\, 2d$

. . . . . $a_1\, +\, a_1\, +\, nd\, +\, 1d\, -\, 2d$

...and then what? :wink:

by **maroonblazer** on Sun Nov 13, 2011 11:43 pm

Ahhh...

$2a_1+nd-1d = 2a_1+(n-1)d$

Thank you!

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