The following sequence gives a series of square numbers,

10% of 10=1,

20% 0f 20=4,

30% of 30=9,

40% of 40=16,........

90% 0f 90=81,

100% 0f 100=100,.....

Is it possible to create a similar series that gives us triangular numbers as the result?

(In reply to

Generalised Parametric Solution by K Sengupta)

We consider the form (a, b) = (nr, (n+1)t), with rt = 50

Since r and t are positive integers, it follows that:

(r, t) = (1,50); (2, 25); (5, 10); (10, 5); (25, 2); (50, 1)

The last value of (r,t), that is (50, 1), can be ignored as we would reach 100% at the very second step.

Now, (r,t) = (1,50) gives: (a, b) = (n, 50(n+1)) yielding Series (i)

(r,t) = (2,25) gives: (a, b) = (2n, 25(n+1)) yielding Series (ii)

(r,t) = (5,10) gives: (a, b) = (5n, 10(n+1)) yielding Series (iii)

(r,t) = (10,5) gives: (a, b) = (10n, 5(n+1)) yielding Series (iv)

(r,t) = (25,2) gives: (a, b) = (25n, 2(n+1)) yielding Series (v)

The Series (I) to (v) are displayed hereunder as follows:

Series(i)

1% of 100 = 1

2% of 150 = 3

3% of 200 = 6

4% of 250 = 10

5% of 300 = 15, and so on

.............

.............

Series (ii)

2% of 50 = 1

4% of 75 = 3

6% of 100 = 6

8% of 125 = 10

10% of 150 = 15, and so on

..............

..............

Series (iii)

5% of 20 = 1

10% of 30 = 3

15% of 40 = 6

20% of 50 = 10

25% of 60 = 15, and so on

.............

.............

Series (iv)

10% of 10 = 1

20% of 15 = 3

30% of 20 = 6

40% of 25 = 10

50% of 30 = 15, and so on

................

................

Series (v)

25% of 4 = 1

50% of 6 = 3

75% of 8 = 6

100% of 10 = 10

(Since by the problem, the series concludes as soon 100 percent is reached)