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Logical Reasoning
Number Series
1223334444112222333333444444441112222223333333334……………………………………what is 2888th no in this sequence
Read Solution (Total 10)
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- Ans is:3
1223334444........
First term=1+2+3+4=10
Second term=2+4+6+8=20
It increase in double 10,20,30,40,60..
In 23rd term sum of total is=2760
2888-2760==128
24th term 1 will cme 24 times,2 will cme 48 times,3 will cme 72 times..
24+48+72=144
So 144 is>128
Ans is:3
- 10 years agoHelpfull: Yes(26) No(2)
- ans is: 3. there are 1 1s 2 2s 3 3s 4 4s . and for the next series it is multiplied by 3 of its numbers and so on.
- 10 years agoHelpfull: Yes(4) No(0)
- 1 to 4 as a set it repeats in order 10,20,30,...... so 23 full sets sums up to 2760 so now in 24th set 1's and 2's sums to 72 and next 72 will be 3's so i guess 3 will be 2888th number
- 10 years agoHelpfull: Yes(3) No(0)
- ANS is 3...
- 10 years agoHelpfull: Yes(1) No(1)
- Sorry the answer will be 3 only...
- 10 years agoHelpfull: Yes(1) No(0)
- Here it form GP with C.R=2
therefore : 1(2^n-1)/(2-1) + 2(2^n-1)/(2-1) + 3(2^n-1)/(2-1) +4(2^n-1)/(2-1) = (2^n-1)10
for n=8
(2^8-1)*10=2560
2888-2560=328.
now the series starts with 9 1's,10 2's,11 3'3 and so on...
hence 2888th term will be 2.
Answer is 2
- 10 years agoHelpfull: Yes(0) No(4)
- 1,2,3,4 then 2,4,6,8 and continues......answer will be 1 at 2888th term
- 10 years agoHelpfull: Yes(0) No(1)
- ans is 3.firstcount digits upto 4 total are 10 next 20....
so using n(n+1)/2*10 - 10 years agoHelpfull: Yes(0) No(0)
- first term has 1 1's 2 2's 3 3's and 4 4's total nos wil be 10
second terms has 2 1's 4 2's 6 3's and 8 4's total nos wil be 20
it goes in such a way like 10+20+40+80+160+320+640+1280=2550 2888-2550=338
remaining there will be 338 digits nineth term has 2^9 1s so ans will be 2 - 10 years agoHelpfull: Yes(0) No(1)
- ans. is 3 but they are not becoming twice
2 is becoming twice ,3 is becoming thrice so
after 23rd
1 is 23 times=23
2 is 2*23times=46
3 is 3*23times=69
- 10 years agoHelpfull: Yes(0) No(0)
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