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 $1000$.
[Math] Arithmetic series question help
sequences-and-series
Related Solutions
The $i^{\text{th}}$ of the first series (if we start counting at zero) is $a+i(b-a)$. The $i^{\text{th}}$ term of your combined series is then $[a+i(b-a)][w+i(x-w)]=aw+i[a(x-w)+w(b-a)]+i^2(b-a)(x-w)$ Now you can use the sum of powers formulas to sum over the range of $i$ you desire.
Suppose that $a_n=a_0+a'n$ and $b_n=b_0+b'n$ are the arithmetic progressions. If there is at most one common term, then obviously it is not an arithmetic progression, and if $a'$ and $b'$ have different signs (one is increasing and one is decreasing), then there are only finitely many common terms so it is not an arithmetic progression. Otherwise, suppose that $a_i=b_k$ and $a_j=b_m$ are the first two common terms, so that $0\le i<j$ and $0\le k<m$ (we can choose $i<j$, and $k<m$ is a consequence of the fact that the sequences are either both increasing or both decreasing).
Now let $c_n=a_i+n(a_j-a_i)$. This is an arithmetic progression, and I claim that all common terms are of this form. To see that every $c_n$ is a common term, note that since
$$a_r+(a_j-a_i)=(a_0+a'r)+(a_0+a'j)-(a_0+a'i)=a_0+a'(r+j-i)=a_{r+j-i},$$
if $a_r=b_s$ is a common term, then so is $a_r+(a_j-a_i)=b_s+(b_m-b_k)$. And then since $c_0$ is a common term, by induction every $c_n$ is.
Now suppose that $a_p=b_q$ is the $p$-minimal common term that is not equal to $c_n$ for any $n$. Since $a_i=b_k$ and $a_j=b_m$ are the first two common terms, which are equal to $c_0$ and $c_1$ by construction, we know $i<j<p$, so $p'=p-(j-i)$ is strictly between $i$ and $p$, and $q'=q-(m-k)$ is between $k$ and $q$. But by the same calculation as above, $a_{p'}=b_{q'}$, so it is a common term, and if it were a $c_n$ for some $n$ then $a_p$ would be $c_{n+1}$, so we conclude that $a_{p'}=b_{q'}$ is another common term not equal to any $c_n$, in contradiction to the minimality of $p$. Thus $c_n$ enumerates all common terms.
Best Answer
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.