Question

Find number of selections of three different numbers from set {1, 2, 3, ...., 101} so that they form an A.P.

Answer 1

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Task. Find number $s$ of selections of three different numbers from set $A=\{1,2,3,....,101\}$ so that they form an ariphmetic progression.

Solution. We should choose numbers of the form $(a,a+d,a+2d)$ for some $a\in A$ and $d>0$. In particular, this implies that

$1\leq a\leq a+2d\leq 101.$

For each $a\in A$ let $s_{a}$ be the number of $d>0$ satisfying the inequaly above: $a+2d\leq 101$. Then

$s=\sum_{a=1}^{101}s_{a}.$

Notice that $a+2d\leq 101$ implies that

$2d\leq 101-a\qquad\Rightarrow\qquad d\leq\frac{101-a}{2}.$

Thus

$1\leq d\leq\frac{101-a}{2},$

and so

$s_{a}=\left[\frac{101-a}{2}\right],$

where $[x]$ is the integer part of $x$, e.g. $[3.5]=3$.

Suppose $a=2k+1$ is odd. Then

$s_{a}=\left[\frac{101-a}{2}\right]=\left[\frac{101-2k-1}{2}\right]=\left[\frac{100-2k}{2}\right]=50-k.$

In particular,

$s_{1}=s_{2*0+1}=50-0=50,$

$s_{3}=s_{2*1+1}=50-1=49,$

$\dots\dots\dots\dots\dots\dots$

$s_{99}=s_{2*49+1}=50-49=1,$

$s_{101}=s_{2*50+1}=50-50=0.$

Recal that

$1+2+\dots+n=\sum_{i=1}^{n}i=\frac{n(n-1)}{2}$

as the sum of first $n$ members of ariphmetic progression $1,2,3,\dots$

Hence

$s_{odd}=\sum_{a\leq 101,\ a\ is\ odd}s_{a}=\sum_{n=0}^{50}k=\frac{50*(50-1)}{2}=1225.$

Now let $a=2k$ be even Then

$s_{a}=\left[\frac{101-a}{2}\right]=\left[\frac{101-2k}{2}\right]=\left[\frac{101-2k}{2}\right]=[50.5-k]=[50-k+0.5]=50-k.$

In particular,

$s_{2}=s_{2*1}=50-1=49,$

$s_{4}=s_{2*2}=50-2=48,$

$\dots\dots\dots\dots\dots\dots$

$s_{98}=s_{2*49}=50-49=1,$

$s_{100}=s_{2*50}=50-50=0.$

Hence

$s_{even}=\sum_{a\leq 100,\ a\ is\ even}s_{a}=\sum_{n=0}^{49}k=\frac{49*(49-1)}{2}=1176.$

Thus

$s=s_{odd}+s_{even}=1225+1176=2401.$