Question

\(\large\tt \begin{align} \color{black}{S={\{1,2.3..........1000}\}
}\end{align}\)
how many arithmatic progressions can be formed from
the elements from \(\Large S\) that start with \(\Large 1\)
and with \(\Large 1000\) and have at least \(\Large 3\)
elements?
a.3
b.4
c.6
d.7
e.8

Answer 1

\(\large\cal\color{red}{Hey}~\color{green}{there}\)

Answer 2

@tom982

Answer 3

Well the bad news is that I haven't covered this before, but the good news is that there's loads of answers for this on Google:

"Let number of terms in progression be n, then using tn = a + (n – 1) d

1000 = 1 + (n – 1)d, or, (n – 1)d = 999 = 33 × 37.

Now, making a table of values of n and d gives us the following values of (n-1):

(n –1) = 3 or 9 or 27 or 37 or 111 or 333 or 999

Hence, 7 progressions are possible."

Source: http://gradestack.com/Complete-CAT-Prep/Consider-the-set-S-1-2-3-1000-/13-3794-3813-19303-37172-sf

Answer 4

thank very much

Answer 5

nice problem