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

Investigation Of The First Form by K Sengupta)

We consider the form (a, b) = (n+1)r, nt), 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 (vi)

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

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

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

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

The Series (vi) to (x) are displayed hereunder as follows:

Series(vi)

2% of 50 = 1

3% of 100 = 3

4% of 150 = 6

5% of 200 = 10

6% of 250 = 15, and so on

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

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

Series (vii)

4% of 25 = 1

6% of 50 = 3

8% of 75 = 6

10% of 100 = 10, and so on

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

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

Series (viii)

10% of 10 = 1

15% of 20 = 3

20% of 30 = 6

25% of 40 = 10, and so on

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

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

Series (ix)

20% of 5 = 1

30% of 10 = 3

40% of 15 = 6

50% of 20 = 10

60% of 25 = 15, and so on

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

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

Series (x)

50% of 2 = 1

75% of 4 = 3

100% of 8 = 6

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

is reached)