# Thread: Arithmetic Sequence-What Am I Doing Wrong?

1. ## Arithmetic Sequence-What Am I Doing Wrong?

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?

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

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

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

4. 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.

5. 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.

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

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

8. ## arithmetric sequence

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