25,36,49
This is a series of perfect squares, 1^2, 2^2, 3^3, 4^2, 5^2....
2006-08-17 15:54:56
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answer #1
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answered by just♪wondering 7
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There are 2 methods of answering this question.
1st method:
1^2=1
2^2=4
3^2=9
4^2=16
The sequence is n^2.
If you need the 5th term, substitute n with 5.
If you need the 6th term, substitute n with 6.
. . . and so on.
5^2=25
6^2=36
7^2=49
So the next 3 terms are 25, 36 & 49.
2nd method:
1+3=4
4+5=9
9+7=16
Each time, the odd numbers, 3, 5, 7, 9. . . are added in sequence to the previous number.
So, you arrive at the next 3 numbers by simply adding 9, 11 & 13 to the previous number.
16+9=25
25+11=36
36+13=49
So, the next 3 numbers in the sequence are 25, 36, 49.
2006-08-17 16:26:54
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answer #2
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answered by Julian 3
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25 36 49
2006-08-17 16:01:16
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answer #3
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answered by Sanket P 2
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1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
ANS : 25, 36, 49
2006-08-17 16:29:02
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answer #4
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answered by Sherman81 6
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16+9= 25
25+11= 36
36+13= 49
2006-08-17 15:55:03
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answer #5
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answered by Kaitelia 5
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Since the given numbers can be written as
(1)^2 , (2)^2 , (3)^2, (4)^2
therefore next three numbers will be
(5)^2 , (6)^2, (7)^2 or 25,36,49
2006-08-17 17:24:39
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answer #6
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answered by Amar Soni 7
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There are really 2 possibilities to this problem. One,you could go for the obvious which are sequenctial 25,36,49 or :your solution could be 27,40,57. If you review your original numbers,you are either adding every other number for the obvious-3,5,7,9-11,13,15 respectively,but if you look at this closer,you will notice that all additions done by you are PRIME NUMBERS-1,3,4,7. So due to the fact that the next primes would be:11,13,17,thus the two possible solutions.
2006-08-17 16:14:31
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answer #7
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answered by dennis p 1
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25, 36, 49 keep on adding the odd numbers
2006-08-17 15:54:53
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answer #8
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answered by cooler 2
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25, 36, 49
2006-08-17 16:55:13
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answer #9
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answered by jimbob 6
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25, 36, 49
2006-08-17 15:58:05
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answer #10
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answered by minhtung91 3
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25, 36, 49
2006-08-17 15:54:10
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answer #11
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answered by Miranda 3
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