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An arithmetic series has first term $5$ and the tenth term is equal to $26$. Find the common difference hence find the least value of $n$ for which the sum of the first $n$ terms of the series exceeds $1000$.

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Welcome to Math.SE! Since you're new here, I'd like mention that it would help if you show what you have already tried. This lets us know where you got stuck, and allows us to better answer your question. Also, showing your work demonstrates that you have invested some time in the problem, which increases the likelihood of getting a helpful answer. –  anorton Apr 2 '13 at 18:42
Hint: Your series is given by $a_n = 5+\frac{7}{3}(n-1),\,n\geq 1$. Then, what would be the sum of the first $k$ terms? –  Lord Soth Apr 2 '13 at 18:48
The common difference is 7/3. I got that by substituting a with 5, n with 10 and 26 with T10, therefore making d, 7/3. –  Heidi Apr 2 '13 at 18:48
Hint2: We have $1+2+\ldots+n = \frac{n(n+1)}{2}$. –  Lord Soth Apr 2 '13 at 18:49

1 Answer 1

Recall that the explicit formula for an arithmetic series is: $$a_n = a_0 + nd \tag{1}$$ ... where $d$ is the common difference.

So, you're given that $a_{9} = 26$ and that $a_0 = 5$. With that information, can you solve equation $(1)$ for $d$?

For the sum question, what you want is the first $n$ such that: $$\sum_{k=0}^{n}a_k \gt 1000 \tag{2}$$

This can be done brute-force with a calculator (or pencil and paper with a lot of patience), or with formulas.

We know that $\sum_{k=0}^{n} k = \frac{n(n+1)}{2}$, and that $\sum_{k=0}^{n}1 = n+1$. Thus, plugging equation $(1)$ in to equation $(2)$: $$\sum_{k=0}^{n} a_0 + kd \gt 1000 $$ $$\sum_{k=0}^{n} a_0 + \sum_{k=0}^{n}kd \gt 1000$$ $$a_0\sum_{k=0}^{n} 1 + d\sum_{k=0}^{n}k \gt 1000$$ $$a_0(n+1) + d\frac{n(n+1)}{2}\gt 1000\tag{3}$$

So, all you need to do is solve $(3)$ for the smallest integer $n$ for which the equality holds.

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If term $1$ is $a_0$ corresponding with n = 0, then term $10$ is $a_9$ corresponding with n = 9. So we have $a_0 = 5$, $a_9 = 26$ as givens. Just wanted to point out the correspondence with 10th term and n = 9. –  amWhy Apr 2 '13 at 19:14
@amWhy Oops. Thanks for pointing that out. For future readers: Make sure you don't do what I just did by forgetting to account for the starting index! :) –  anorton Apr 2 '13 at 20:28
Now it is indeed worth my +1! Good work! –  amWhy Apr 2 '13 at 20:34

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