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6
Alternating numbers are going in series: 1 2 3....
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Ans is 6
there are many series embedded in each other
in first there is a gap of one term between two consecutive terms
in second there is gap of three terms between two consecutive terms
in third there is gap of seven terms between two consecutive terms
in third series last term is 5 and after a gap of seven term x=6
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so removing those series numbers we get a sequence in which we have another series in alternative sequence(index number 2, 6, 10 and so on). Following the series, we get the number.
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
1 1 2 1 3 2 4 1 5 3 6 2 7 4 8 1 9 5 10 3 11;
so on... the alternate numbers form increasing series.. so answer is 6.
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x = 6
Logic:
The sequence in itself has multiple sequences alternating in between.
1, seq2, 2, seq3, 3, seq2, 4, seq3, 5,..
seq2 is again 1,2,3,4,5..
And seq3 is half the value of the 1st even number happening to it's right side
So at the end we have ..22, seq3, 23, seq2
Here the seq2 number is 12
So x=12/2 = 6
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removing the first set of alternate terms which ar 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 we get
1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, X, ...
1, 1, 2, 1, 3, 2, 4, 1, 5, 3, X, ...
1, 1, 2, 1, 3, X, ... so 6 miust come here
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Check this out. Of course, this does not mean that I am the winner. ET, RVignesh, Varsha and CP had answered it correctly before me today.
Congos ET, CP, RVignesh and Varsha. 😀
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1, 1, 2, 1,
3, 2, 4, 1,
5, 3, 6, 2,
7, 4, 8, 1,
9, 5, 10, 3,
11, 6, 12, 2,
13, 7, 14, 4,
15, 8, 16, 1,
17, 9, 18, 5,
19, 10, 2, 3,
21, 11, 22, x
1[SUP]st[/SUP] Sequence: 2n-1 (1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21)
2[SUP]nd[/SUP] Sequence: n (1, 2, 3,…., 21)
3[SUP]rd[/SUP] Sequence: 2n (2, 4, 6,……, 22)
4[SUP]th[/SUP] Sequence: 1, 1, 2, 1, 3, 2, 4, 1, 5, 3, x
Here again:
1, 1, 2, 1,
3, 2, 4, 1,
5, 3, 6, 2….and the same sequence continues.
So answer is 6.
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Answer 6
in the given series we can see 2 sequences.. one is 1,2,3,4.....,22
other is 1,1,2,1,3,2,4,1.......,11 (alternating numbers of the given sequence)
the 2nd sequence above has again 2 sequences..
one is 1,2,3,4....., 11
other is 1,1,2,1,3,2,4,1,5,3,..
it contains 2 sequences again..(1,2,3,4,5 and 1,1,2,1,3)
and the missing number is the next number in the sequence 1,2,3,4,5..
that is, 6
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the answer is 6
the alternate series is going on...
in every series between every consecutive term there is no 1
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brain
Member •
Nov 5, 2011
its 6...after 4 dgts whn 1 coms it added wd 1(frst tym),next tym 1 is added wd 1(1+1=2),,next 2 is added wd 2(2+2=4),,at last 3 is added wd 3(3+3=6)............thas all the game!
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X=6
consider the series as
a,b,c,d,e,f,g,...
4th number is a,
8th number is b,
12th number is c,
16th number is d...
So,
40th number is k
(i.e., k=6).
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Answer is 6
Bcoz every number is coming after certain period like
1 is on position 1 then 2,4,8,16
2 is on 3,6,12.......
and x is on 44th position so half of 44 is 22 th position where is 6...
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