Be Reasonable

Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.

In y = ax +b when are a, -b/a, b in arithmetic progression. The polynomial y = ax^2 + bx + c has roots r1 and r2. Can a, r1, b, r2 and c be in arithmetic progression?

Proof Sorter - Sum of an AP

Use this interactivity to sort out the steps of the proof of the formula for the sum of an arithmetic series. The 'thermometer' will tell you how you are doing

Summats Clear

Stage: 5 Challenge Level:

Vassil from Lawnswood School, Leeds, Michael from Madras College St Andrews and Koopa Koo from Boston College all solved this problem, well done all of you.

Here is Vassil's solution:

Let $f(n)$ denote the sum of the first $n$ terms of the sequence $$0, 1, 1, 2, 2, 3, 3,\ldots , p, p, p+1, p+1,\ldots.$$

First I tried with several numbers. Let $n=15$. Then $f(15)=2 \times (1+2+3+4+5+6+7)=7 \times 8$
where the sequence is: 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7.

Let $n=14$. Then $f(14)=2 \times (1+2+3+4+5+6+7)-7=7 \times 7$
where the sequence is: $0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7$.

Let $n=17$. Then $f(17)=2 \times (1+2+3+4+5+6+7+8)=8 \times 9$
where the sequence is: $0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8$.

Let $n=16$. Then $f(16)=2 \times (1+2+3+4+5+6+7+8)-8=8 \times 8$
where the sequence is: $0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8$.

I noticed that the formula for $f(n)$ depends on whether $n$ is odd or even.

${\bf Case}$ ${\bf I}$ - $n$ is odd, i.e. $n=2k+1$

Then \eqalign { f(n) = 2(1+2+...+k)\cr = 2k(k+1)/2 \cr = {\left(n - 1\over 2\right)}{\left(n+1\over 2\right)}.}

${\bf Case}$ ${\bf II}$ - $n$ is even, i.e. $n=2k$

\eqalign { f(n) = 2(1+2+...+k) - k \cr = k^2 + k - k \cr = k^2 \cr = \left({n\over 2}\right)^2.}

Now we have to calculate $f(a+b)-f(a-b)$.

There are two cases. In the first case, when one of $a$ and $b$ is even and the other is odd, then $(a+b)$ and $(a-b)$ are both odd. Otherwise $(a+b)$ and $(a-b)$ are both even.

Case I $(a+b)$ and $(a-b)$ both odd. \eqalign{ f(a + b) - f(a - b) = {(a + b)^2 - 1\over 4} - {(a - b)^2 - 1\over 4} \cr = ab.} Case II $(a+b)$ and $(a-b)$ both even. \eqalign{ f(a + b) - f(a - b) = {(a + b)^2\over 4} - {(a - b)^2\over 4} \cr = ab.}