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All I know is that a=13 and common difference=8. I am unaware how to continue from here.

Melonie Seltzer
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  • The answers you've gotten are helpful. If it's still a bridge too far to understand them, though, the numbers here are small enough that you could just add them until you exceed $1,000$. The answer's less than $20$ ... – John Sep 12 '17 at 23:28
  • I know from the back of my math book that the answer is 15. My problem is that I don't know a formula to get there. I appreciate the equations I've gotten but I am simply not at a level of math that has even introduced them to me yet. – Melonie Seltzer Sep 12 '17 at 23:36
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    Here's an excercise to do on your own time. Write on a piece of paper $1 + 2 + 3 + 4 + 5 + .............+ 99 + 100$. Then try adding them up by adding the $1$ and the $100$. Then add the $2$ and the $99$. Then add the $3$ and $98$. And so on. What do you get? Then do it again with $1 + 2 + 3 + ....+ 253$. Then do it for $1+2+3+ .... + n$. After you do that exercise go back to this problem. See if you have gained any insight. – fleablood Sep 13 '17 at 00:09
  • Any time you don't know "how to continue from here" on a sequence of integers, look it up in the OEIS. 13,21,29,37 gives just fourteen results. One of them is bound to be what you're looking for. – Robert Soupe Sep 13 '17 at 18:28

5 Answers5

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The sum of an arithmetic sequence with initial term $a$ and common difference $d$ can be stated as

$$S_n = a + (a + d) + (a + 2d) + (a + 3d) + ... + (a + (n-1)d)$$

Or

$$\begin{align} S_n &= \sum_{k=0}^{n-1}(a + kd)\\ & = \sum_{k=0}^{n-1}a + \sum_{k=0}^{n-1}kd \\ & = a \cdot \sum_{k=0}^{n-1}(1) + d \cdot \sum_{k=0}^{n-1}k \\ & = an + d \cdot \sum_{k=0}^{n-1}k \\ & = an + d \cdot \left(\frac{n(n-1)}{2}\right) \\ \end{align}$$

Can you take it from here?

Marcus Andrews
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Hint: Can you represent the $n$th term as a function of $n$? From there, can you use that function to find the sum of the first $n$ terms as a function of $n$? From there, can you figure out when that first eclipses $1000$?

Carl Schildkraut
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  • I really just don't know what I'm doing. If someone could give me a straight answer that would be great. – Melonie Seltzer Sep 12 '17 at 23:06
  • @MelonieSeltzer This isn't a website for people to give you answers to homework problems. Do you know how to represent the $n$th term as a function of $n$? – Carl Schildkraut Sep 12 '17 at 23:09
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The constant difference of $8$ suggests this representation will be useful:

$$ a_1 = 13 = 5+ 8\times 1 \\ a_2 = 21 = 5+ 8\times 2 \\ a_3 = 29 = 5+ 8\times 3 \\ a_4 = 37 = 5+ 8\times 4 \\ $$

Then the sum looks like this:

$s_k = \sum_{i=1}^k(5+8i) = 5k + \sum_{i=1}^k(8i) = 5k + 8\cdot\sum_{i=1}^k i$

And that sum is the triangular numbers, with the fairly well-known formula

$\sum_{i=1}^k i = k(k+1)/2$

So you are looking for when $s_k = 5k+4k(k+1) = 4k^2+9k$ goes past $1000$. And since $16^2>250$, you won't have far to look.

Or, if you prefer, you can solve the quadratic $4k^2+9k-1000=0$

John
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Joffan
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Just to do it slightly differently.

If I have $a_0 = 13$ and $a_k = 13 + 8*k$ and I have a list of $n$ of these terms $a_0=13, a_1=13+8, a_2=13+2*8,......, a_n=13 + n*8$, what is the average value of all of them.

Well as they are all exactly $8$ apart from each other, the average value will be directly in the middle or $13 + \frac n2*8$.

So what do you get when you add up all the values? Well you nave $(n+1)$ terms. And the average term is $13+\frac n2*8$. So when you add them all up you should get $(n+1)(13 + \frac n2*8)$.

So $(n+1)(13+\frac n2*8) > 1000$

....

But one thing you should teach yourself. $1 + 2 + 3 + 4 + 5 + ...... +n=???$.

Well the average is $\frac {n+1}2$. ANd there are $n$ terms so $???? = \frac {n(n+1)}2$. You will see that formula a lot!

Alternative proof:

$1 + 2 + 3 + 4 + .......... + n = S$

$n + (n-1) + (n-2) + (n-3) + ... + 1 = S$

$(n+1) + (n-1+2) + (n-2+3) + (n-3 + 4) + ..........+(1+n) = 2S$

$(n+1) + (n+1)+ (n+1) + (n+1) + ...... + (1+n) = 2S$

$n(n+1) = 2S$

$\frac {n(n+1)}2 = S$.

.....

So we have $13 + (13 + 8) + (13+2*8) + (13+2*8) + ...... + (13+n*8) = $

$[13 + 13 + 13 + ....... + ] + [8 + 2*8 + 3*8 + ........ + n*8]=$

$(n+1)13 + 8[1 + 2 + 3 + ........ + n] =$

$(n+1)13 + 8*\frac {n(n+1)}2 = $

$(n+1)(13 + 4n) > 1000$.

fleablood
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Let's look a bit more at Marcus's expressions:

$$ a_1 = 13 = 5+ 8\times 1 \\ a_2 = 21 = 5+ 8\times 2 \\ a_3 = 29 = 5+ 8\times 3 \\ a_4 = 37 = 5+ 8\times 4 \\ ... \\ a_n = 5 + 8\times n \\ $$

What we see from this is that if we add up all $n$ of the terms (call this $S_n$) we have two parts:

  1. $n$ fives, whose sum is $5n$
  2. $8$ times the sum of the first $n$ positive integers

The second part has a formula that may have been in your textbook: $n(n+1)/2.$ As an example: $1+2+3+4+5 = 5(5+1)/2 = 30/2 = 15.$

So,

$$S_n = 5n + \frac{8n(n+1)}{2}.$$

This is an expression for the sum of the first $n$ terms in your sequence. It's specific to this sequence. The other answers give formulas that can be applied to different sequences, but this one works for your sequence.

Here, if you want, you can try values of $n$ to see what the sum is. Eventually, you'll find that $n=14$ gives a sum less than $1000$, and $n=15$ gives a sum greater than $1000$. So, the answer is $15$.

But you can do it without trial and error, too. We can set the sum to $1000$:

$$1000 = 5n + \frac{8n(n+1)}{2}.$$

Now, we need to solve for $n$. To do this we need to put this into a form that we can solve:

$$2000 = 10n + 8n^2 + 8n$$ $$8n^2 + 18n - 2000 = 0.$$

We can solve this with the quadratic formula:

$$n = \frac{-18 \pm \sqrt{18^2 - 4(8)(-2000)}}{2(8)}.$$

Plug this into a calculator, and you'll get a negative number (which is clearly not the right answer) and a number a bit more than $14$. So we need at least $15$ terms to reach $1000$.

John
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