The arithmetic sequence is 13, 21, 29, 37... Find the least number of terms required for the sum of the sequence terms to exceed 1000. All I know is that a=13 and common difference=8. I am unaware how to continue from here.
 A: 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?
A: 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 $1000$?
A: 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 $4k^2+9k-1000=0$ 
A: 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 = $
$(n+1)(13 + 4n) > 1000$.
A: 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:


*

*$n$ fives, whose sum is $5n$

*$8$ 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 $1000$.
