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(#M40012002) TCS Logical Reasoning question Keep an EYE

1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,3,3,3,3,4,4,4,4,4..........................what would be 2320th element of this series??

Asked In TCS manish mohan (11 years ago)

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Read Solution (20) :

Pradeep  (11 years ago)

Answer=> 2320th element is 1.
Solution:
set 1= 1,2,2,3,3,3,4,4,4,4 =10 elements
set 2= 1,1,2,2,2,3,3,3,3,4,4,4,4,4 = 14 elements
therefore
10+14+.......+n-1+n+M=2320
set1+set2+....setn+M=2320
[M included because we dont know that 2320 is an exact sum of this A.P series]
set1+set2+...set16=2304
so M=16
After set 16, set 17 has 17 ones. So M=16 so 2320th element is "1".

Is this solution Helpfull? Yes (13) | No (3)  

Rahul singh  (11 years ago)

1 will be ans

Is this solution Helpfull? Yes (8) | No (5)  

manish mohan  (11 years ago)

can anyone explain properly???? in exam they are changing the value of element.. so ometime ans is one, and sometime it is four,two ,three... we cant just learn...

Is this solution Helpfull? Yes (7) | No   

Arka Laha  (11 years ago)

we can see in first 10 terms there is one 1,then there will be 14 terms where two 1s next in 18 there are three 1s....
so i have to do first, n/2((2*10)+(n-1)*4))2n^2 + 8n2(32)^2 + 8*32=2304 n=32 where this is satisfying and the 32nd term of(10,14,18..)will be 134 and in 134 there are 32 ones and in the next term there will be 33 ones 34 two but after 134 the next (2320-2304)=16 th term is 2320th term and there will be a one becoz in 138 there is 33 ones....... This is the one thing i am trying to explain this is the correct way according to me.....

Is this solution Helpfull? Yes (6) | No (3)  

niki  (11 years ago)

please some1 explain properly how to solve such problems..

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semanti dey  (11 years ago)

ans is 4..

Is this solution Helpfull? Yes (2) | No (6)  

Supratik Chakraborty  (11 years ago)

after the 2304th element there will be 32 1s. so the 2320th element must be 1

Is this solution Helpfull? Yes (1) | No   

Arka Laha  (11 years ago)

the terms are increasing by 3 like 10,13,16 like this...
10+13+16+....+115=2250,in ten there is one 1,in 13,two 1 like this in 118 there is 36 1,37 2...so as 2320-2250=70

Is this solution Helpfull? Yes (1) | No (3)  

Arka Laha  (11 years ago)

sorry in 118 there will be 37 ones nd 38 twos, it is also(37+38)=75>70,so ans is 2...

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Deep  (9 years ago)

as we all know the international formula of A.P ...
Sn=n/2[2a+(n-1)d]
so here on first set 10,second set 14 and must be the third 18 so d=4.
a=10,we just need to find out the n ;
as here
n/2[2a+(n-1)d]

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gopal kumar  (11 years ago)

4 is the correct answer

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gopal kumar  (11 years ago)

send me recent tcs paper at kumar.gopal308@gmail.com pleae

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KD  (11 years ago)

1 sure........

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Arka Laha  (11 years ago)

sorry frnds the common diff is 4 i hae mistaken with 3,so it will be 10,14,18,........134 and sum of them is 2276 and 2320-2304=16 in 10 there is one 1 in 14 there is two 1.... so in 138 there will be 33 ones which is greater than 16 so ans is one... plz forgive for the previous two wrong ans......

Is this solution Helpfull? Yes | No (1)  

sharmi mukherjee  (11 years ago)

the 2320th term of the series is 2 which will be 732 times

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semanti dey  (11 years ago)

yes its 1..my previous ans ws wrong..

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shashank  (11 years ago)

surely ans. is 1....

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Deep  (9 years ago)

as we all know the international formula of A.P ...
Sn=n/2[2a+(n-1)d]
so here on first set 10,second set 14 and must be the third 18 so d=4.
a=10,we just need to find out the n
as here
n/2[2a+(n-1)d]

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rajesh  (8 years ago)

In first term 1,2,2,3,3,3,4,4,4,4 then count=10
In second term count is 20
next is 30 and so on.
so using n(n+1)/2 *10

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pritam  (3 years ago)

in first set: 1,2,2,3,3,3,4,4,4,4 total terms=10 terms and next set will start with value '1'
2 set: 1,1,2,2,,2,2,3,3,3,3,3,3,4,4,4,,4,4,4,4,,4=20 terms and next set will start with value '1'
3 set: each term is 3 times of its value ,e.g= 2*3 ie 2 will be 6 times ,therefore in all =30 terms

similarly
231^th set will have 2310 terms and next set will start with '1'
now in next set which is 232 will have value '1' 232 times i.e from 2311 to (2310+232)i.e upto 2543 term value will remain 1 ,and we need 2320 term which will also be '1'

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