The sum of K consecutive integers is 41. If the least integer is -40, then k?

- anonymous

The sum of K consecutive integers is 41. If the least integer is -40, then k?

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- schrodinger

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- anonymous

Here's a way of thinking about this. If you're adding consecutive integers and the first is -41, then you're adding:\[-41+(-40)+(-39)+(-38)+...=41\]However, at some point, you're going to get into the positive numbers when you're adding.\[-41+(-40)+(-39)+...+(-3)+(-2)+(-1)+(0)+(1)+(2)+(3)+...=41\]At the point where you get to the positive numbers, the negatives and positives will begin to cancel out. This is because -3 + 3 = 0, -2 + 2 = 0, -1 + 1 = 0. Does this help you with the rest of the problem?

- anonymous

The cutoff part says \(+...=41\)

- phi

the least number is -40

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## More answers

- misty1212

oops

- anonymous

Ok, if the first numebr is -40, just omit the -41 in my equations.

- dan815

does that equation work

- anonymous

hmmmmm so it's 41 then?

- anonymous

Not quite. 41 would be the last term that you would be adding to get the sum to be 41. However, k is the number of numbers you're adding.

- anonymous

ohhhhh crap .... it's 81

- anonymous

-40 to 0 is 41 numbers. 1 to 41 is 41 numbers. So, 41 + 41 should be what k is. Is this making sense to you?

- freckles

\[\sum_{i=1}^k(-40+i)=41 \\ -40k+\frac{k(k+1)}{2}=41\]

- freckles

it is a quadratic equation

- anonymous

hmmm you know what i do is i add 41 to -40 and then multiply by 81

- freckles

I think I should have started it at i=0

- anonymous

The key here is that they are "consecutive" integers. That's why just multiplying like that doesn't work.

- freckles

\[-40(k+1)+\frac{k(k+1)}{2}=41\]

- anonymous

crap so it's 82

- freckles

almost

- anonymous

huh?

- anonymous

@freckles I got 82 as well. Might've had an error though.

- freckles

you said 81 earlier that is right

- freckles

that is what you get when you solve that quadratic equation above
\[\sum_{i=0}^k (-40+i)\]
this does mean
-40+(-39)+(-38)+...+(-40+k)
or if you used those formula things
you can write that sum thingy as
\[-40(k+1)+\frac{k(k+1)}{2} \text{ which we want to this output } 41 \\ \text{ so we need \to solve } \\ -40(k+1)+\frac{k(k+1)}{2}=41\]

- freckles

\[\frac{-80(k+1)}{2}+\frac{k(k+1)}{2}=41 \\ \frac{(k+1)}{2}(-80+k)=41 \\ (k+1)(-80+k)=82 \\ k^2-79k-80=82 \\ k^2-79k-80-82=0 \\ k^2-79k-162=0 \\ (k+2)(k-81)=0\]

- freckles

only one of these k's make sense

- dan815

82 is right i believe

- dan815

|dw:1436823495199:dw|

- anonymous

@freckles You can use the sum of an arithmetic series.\[a_k=-40, -39, ...\]\[S_k=\frac{k}{2}(2a_1+(k-1)d)\]\[41=\frac{k}{2}(k-81)\]\[82=k^2-81k\]\[0=k^2-81k-82\]\[0=(k-82)(k+1)\]\[k=-1 ~or ~82\]And -1 is extraneous.

- anonymous

If you prefer the algebraic method.

- freckles

I gave an algebraic method above
81 works

- freckles

http://www.wolframalpha.com/input/?i=sum%28-40%2Bi%2Ci%3D0..k%29%3D41

- anonymous

By starting at i = 0, the number of terms in that series is actually k + 1. I think that's the error.

- anonymous

thanks everyone.

- dan815

yeah 82 is right

- freckles

then why is wolfram saying 81 that is weird

- dan815

freckles the indice starts at 0

- dan815

not 1

- anonymous

The number of terms can easily be calculated by b-a+1 given\[\sum_{a}^{b}\]What you used equates to k+1 terms.

- dan815

|dw:1436823854020:dw|

- freckles

so I should have used this:
\[\sum_{i=1}^{k}(-40+(i-1)) =41 \\ -40k+\frac{k(k+1)}{2}-k =41 \]
i think this should arrive to the correct answer
and yes this is k terms now
my bad

- freckles

\[-80k+k(k+1)-2k=82 \\ k^2+k-82k=82 \\ k^2-81k-82=0 \\ (k-82)(k+1)=0\]

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