You may also like

problem icon

Be Reasonable

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

problem icon

Janusz Asked

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?

problem icon

Prime AP

What can you say about the common difference of an AP where every term is prime?

Summats Clear

Stage: 5 Challenge Level: Challenge Level:1

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. $$\eqalign{ f(a + b) - f(a - b) = {(a + b)^2\over 4} - {(a - b)^2\over 4} \cr = ab.}$$