Latest TCS Aptitude Question SOLUTION: What is the 2532nd position of the number in the sequence 1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4, 4,4,4,4,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4, 4,4,4,4,4,4,4,4,

Ans will be 1..
10,20,30,40...
10(1+2+3----+22)=2530
That mean upto 22 th pos total numbers are 2530
2532 th numner must be 1..

1,2,2,3,3,3,4,4,4,4-there are 10 numbers.In the next case there are 20 numbers,then in the next case there are 30 numbers.So each time there is an increment of 10 numbers.So at the 2528th position the number is 4.So at the 2532nd position,the number is 1.

Ans:1

from given series
terms like 1(10), 2(10),3(10).......n(10) then
no.of terms x=1(10)+2(10)+3(10)........+n(10)=10(1+2+3+......+n)=10[(1/2)*n(n+1)]
for every "10[(1/2)*n(n+1)]" th term ends with 4 and next term starts with 1
if n=22 =>x=2530 => 2530 th term ends with 4 and next term starts with 1
but find 2532 , so 2532-2530=2
=.2530 th after 2 terms is 2532=>so 2532 th number is 1
=> 1 is ans

First if we count 1223334444. they are 10
In the next term they are 20
Next they are 30 and so on
So Using n(n+1)2×10≤2532

For n = 22 we get LHS as 2530. Remaining terms 2.

Now in the 23th term, we have 23 1's, and next 46 terms are 2's. So next 72 terms are 3's.
so the 2532 term will be “1”.

4..
there are 1,2,2,3,3,3,4,4,4,4. total 10 numbers.& they are getting double.
so 10,20,40,80,160,320,640,1280=2550.
last numbers are 4...

1 is right answe

if we see the sequence clearly, then we get something like this
(#M40115143) TCS QUESTION KEEP AN EYE
IMAGE: HTTP://WWW.M4MATHS.COM/IMAGES/EYE_CLOSE.PNG

Keep an eye puzzle
What is the 2532nd position of the number
in the sequence
1,2,2,3,3,3,4,4,4,4, (contains 10 terms) (first row)
1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4, (contains 20 terms) (second row)
1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4 (contains 30 terms) (third row)
and so on. So we have to find the number of row in which 2532nd term exists.
so to find the row we have a series like 10, 20, 30, 40, .....
Let nth term gives the row which contains 2532nd term
so n/2 [(2 * 10) + (n-1) *10 ] = 2530 (now why we took 2530 instead of 2532, it is because of easy calculation. From the above equation we will get the value of 'n' as 22.
It means till 22th term 2530 term of the given sequence was being used. So 2531st term will go to the 23rd term and 2532nd term will be the second term of the 23rd row.
So now we have to find the second term of 23rd row. As from the given sequence, we can see that each row contains 1 equal to the number of that row. so 23rd row will contains first 23 terms as 1. So the 2nd terms will be 1.
And 2532nd term of the given sequence will be 1.
NOTE: I am not able to see the options. Can anyone please tell me why?

The given sequence is -> 1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,
4,4,4,4,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4,
4,4,4,4,4,4,4,4,…..

On summing up all equal consecutive numbers,

This can be written as -> 1,4,9,16,2,8,18,32,3,12,27,48,.....

So, here we can see this as separate 4 AP series i.e.,
series 1-> 1,2,3,4,...
series 2-> 4,8,12,16,..
series 3-> 9,18,27,36,..
series 4-> 16,32,48,...

Now, divide the asked position by 4, if we get remainder 3 then answer will be 4
if we get remainder 2 then answer will be 3
if we get remainder 1 then answer will be 2
if we get remainder 0 then answer will be 1

As here position asked is 2532, so we divide 2532 by 4 remainder is 0, hence answer is 1.

2532=12*211.
In the series if we add no. of terms then 211th term is 1.
Therefore n*211 implies again 1

ans to this q. is 1 because...
1. go up to 26th series
2. 25th series ends with 2500
3. now add 2500 +26=2526 and 2500+52=2552(because of 26th series 26 52 78 104....)
4. 2532 comes between 2526 and 2552 thas why ...ans is 1

ans will be 4.
In first sequence their is 10 term.Next sequence have 20 term.10 20 40 80 160 320 640 1280 so the total term is 5540.In the last sequence have 8 no of 1,16 no of 2 & 64 no of 4. So the no will be 4.