Arithmetic Sequence-What Am I Doing Wrong?

**Tazo** · April 14th 2010, 08:11 PM

Even if i try simplifying the fraction it doesn't come out to 65/2, which is the correct answer in the back of the book.

Am i missing something?

**maddas** · April 14th 2010, 08:19 PM

$\sum_{k=0}^{19} {k-3\over4} = \frac14(\sum_{k=0}^{19} k - \sum_{k=0}^{19} 3) = \frac14({19(19+1)\over2} - 20\cdot3) = \frac{65}2$ .

Are you using that $\sum_{k=0}^N k = {N(N+1)\over2}$ ?

**sa-ri-ga-ma** · April 14th 2010, 08:20 PM

a(19) = a(1) + (19-1)d
Now find a(19)

**Tazo** · April 14th 2010, 08:35 PM

Originally Posted by sa-ri-ga-ma

a(19) = a(1) + (19-1)d
Now find a(19)

Already tried that dude. I still got the wrong answer.

Originally Posted by maddas

$\sum_{k=0}^{19} {k-3\over4} = \frac14(\sum_{k=0}^{19} k - \sum_{k=0}^{19} 3) = \frac14({19(19+1)\over2} - 20\cdot3) = \frac{65}2$ .

Are you using that $\sum_{k=0}^N k = {N(N+1)\over2}$ ?

Man, that totally confused me.

I know I've seen that formula before, but i can't seem to remember it.

**sa-ri-ga-ma** · April 14th 2010, 08:49 PM

a(19) = -3/4 + 18*1/4 = 15/4.
So sum is given by
19/2*(-3/4 + 15/4)
Simplify and check the answer.

**Tazo** · April 14th 2010, 09:10 PM

It's still wrong. I get 57/2.

**HallsofIvy** · April 14th 2010, 10:19 PM

Since your sum starts with $a_0$ , not $a_1$ , the number of terms is 20, not 19.

**bjhopper** · April 15th 2010, 12:10 PM

Originally Posted by Tazo

Even if i try simplifying the fraction it doesn't come out to 65/2, which is the correct answer in the back of the book.

Am i missing something?

If the first term is -3/4,common diff 1/4 the sum of 20 terms is 65/2. You should write your questions exactly as in your book. Otherwise everyone is confused.

bjh

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