2006-11-03 18:47:39 · 21 answers · asked by SREE 2 in Science & Mathematics ➔ Mathematics
2A11
4D13
12G17
48J25
240M41
2006-11-03 18:52:03 · answer #1 · answered by Gargi 1 · 1⤊ 0⤋
48 K 22-96 O 29?
2006-11-04 03:02:03 · answer #2 · answered by wheeliebin 6 · 0⤊ 0⤋
i broke them into 3 parts. 1st No., letter, and 2nd No.
2 *2 = 4 *3 = 12
A bc D ef G
11 +2= 13 +4= 17
so the next 2 terms MIGHT BE
48J23
240M31
OR
48J25
240M41
(cos for last no., u add 2, dn 2*2, dn the third n fourth term might be 2*2*2 n 2*2*2*2)
2006-11-04 03:01:39 · answer #3 · answered by luv_phy 3 · 1⤊ 0⤋
First term is being multiplied by the next consecutive number.
2nd term has diff. of 2 each time
Last digits are consecutive primes after 11
therefore 48J19, 240M23
2006-11-04 05:03:18 · answer #4 · answered by sushant 3 · 0⤊ 0⤋
48J25, 240M41
its because
2A(1)11,
2*2(4)D(4)13(11+2),
4*3(12)G(7(4+3))17(13+4),
12*4(48)J(10(7+3)25(17+8),
48*5(240)M(13(10+3))41(25+16)
2006-11-04 03:38:33 · answer #5 · answered by bubbly 2 · 0⤊ 0⤋
2*2=4 4*3=12...
2+11=13 4+13=17...
(A)BC(D)EF(G)...
next numbers:
48J29 , 240M77
2006-11-04 06:30:35 · answer #6 · answered by George 1 · 0⤊ 0⤋
One possibility.
2A11
4D13
12G17
32J29
88M61
2006-11-04 03:23:18 · answer #7 · answered by Brenmore 5 · 0⤊ 0⤋
48J23 and 240M31
or
48J25 and 240M41
2006-11-04 03:22:01 · answer #8 · answered by CJ 3 · 0⤊ 0⤋
32J19 88M23
2006-11-04 09:42:42 · answer #9 · answered by Atul I 2 · 0⤊ 0⤋
24J23,40M31.
choose me the best if it's rite.
2006-11-04 02:59:59 · answer #10 · answered by Anonymous · 0⤊ 1⤋
16f12,19r23
2006-11-04 06:31:20 · answer #11 · answered by venkata reddy s 1 · 0⤊ 0⤋