__,__,__,48,100,180,
294,448,648,__,__,__,.......Find the three numbers before and after given numbers in the series.

(In reply to

Answer by K Sengupta)

Let us denote the pth term of the given sequence by S(p).

It is given that:

S(4) = 48, and we note that 48 divided by the square of 4 is 3, or alternatively 48/(4^2) = 3

S(5) = 100, and we note that 100/(5^2) = 4 = 5-1

S(6) = 180, and we note that 180/(6^2) = 5 = 6-1

and, so on:

Finally, S(6) = 648, and we note that 648/(9^2) = 8 = 9-1

Accordingly, it follows that in general

S(p)/(p^2) = p-1, giving:

S(p) = p^2*(p-1) .....(#)

Thus, putting p = 1, 2, 3, 10, 11, 12 in turn in (#), we obtain:

S(1) = 1^2*(1-1) = 0

S(2) = 4

S(3) = 18

S(10) = 900

S(11) = 1210

S(12) = 1584

Consequently, the first three missing numbers are 0, 4 and 18, while the last three missing numbers are 900, 1210 and 1584.

*Edited on ***February 25, 2009, 11:54 pm**