Latest TCS Aptitude Question SOLUTION: IN THIS SEQUENCE 1 22 333 4444 11 2222 333333 44444444 111 222222 ........... what is 2926th term???

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IN THIS SEQUENCE 1 22 333 4444 11 2222 333333 44444444 111 222222 ........... what is 2926th term???

Asked In TCS Arun Sairam (11 years ago)

Unsolved Read Solution (11)

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nirvan gupta  (11 years ago)

given sequence is of interval 4
2926/4 = results 731 and remainder is 2
i.e 731 sequences of group 4 is completed
reminder 2 tell that next term is 732 times 1 and 1464 times 2
so answer is
2 occuring 1464 times

Is this solution Helpfull? Yes (29) | No (4)  

redlapalli  (11 years ago)

1 2 3 4 = 10
2 4 6 8 = 20
10+20+30+40......=2926
10(1+2+3+4+...)=2926
10(n(N+1)/2)=1296
n = 23.03
2926th term is 4

Is this solution Helpfull? Yes (16) | No (26)  

Amit Mittal  (11 years ago)

2 occurs 1464 times
2926mod4= 2 so 732th block will start
in every block nu,ber of 2's are incremented by 2 so 732*2= 1464 2's

Is this solution Helpfull? Yes (9) | No (1)  

chirag  (11 years ago)

2 occur 732 times is d answer

Is this solution Helpfull? Yes (5) | No (9)  

happy das  (8 years ago)

IN THIS SEQUENCE 1 22 333 4444 11 2222 333333 44444444 111 222222 ……….. what is 2926th term

1 22 333 4444 has 10 terms; 11 2222 333333 44444444 has 20 terms; Like this we will have 30, 40, 50 terms…Please note that in the first 10 terms, the last four terms are all ‘4’, and in the next 20 terms the last eight terms are all ‘4’ .

Let 3000 be the sum of 10. 20, 30,…terms. The number of terms that will give this total of 3000 is given by the formula: n[2a +(n-1)d]/2. We have 3000 = n[20 + (n-1)10] /2 Or 6000 = n(10n +10) or 600 = n(n+1); or n = 24. In other words, if 10 is the first term, 20 is the second term then the 24th term will be 240.

We see that in 10 terms there are four ‘4’; hence in 240 terms there will be ninety six ‘4’. Hence from 3000-96 till 96 we will have nothing but ‘4’s only. i.e, from the 2904th term, we have nothing but ‘4’ only. Hence the 2926th term is 4.

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shrawan kumar  (11 years ago)

222222...upto 1462term

Is this solution Helpfull? Yes (4) | No (2)  

Thrinadh  (8 years ago)

(1 22 333 4444) (11 2222 333333 44444444)
10 + 20 + 30 +40 - - - - - - - - - - - - -consider (290)
the number of terms in the given question are in AP,so the a+(n-1)d = 10+(290-1)10 =2900
for 10 we have 1 1's
for 20 we have 2 1's
like wise for 290 we have 29 1's
and next term will be have 30 1's
so 2900 ends with 4
now upto 2930 all digits are 1's
and 2926th term also will be 1
hope you understood...

Is this solution Helpfull? Yes (2) | No   

lala  (11 years ago)

see the given sequence is after interval 4
2926/4 = results 731 and reminder is 2
i.e 731 sequences of group 4 is completed
reminder 2 tell that next term is 732 times 1 and 732 times 2
so answer is
2 occur 732 times is d answer

Is this solution Helpfull? Yes (1) | No (7)  

Natarajan. R  (8 years ago)

answer is 2. bcoz 732 times 1 and 1464 times is 2

Is this solution Helpfull? Yes (1) | No   

Arun Sairam  (11 years ago)

can u give me a detailed explanation plz.

Is this solution Helpfull? Yes | No (3)  

sowmya  (8 years ago)

total grandtotal
1 2 3 4 10 10
2 4 6 8 20 30
4 8 12 16 40 70
8 16 24 32 80 150
16 32 48 64 160 310
32 64 96 128 320 630
64 128 192 256 640 1270
128 256 384 512 1280 2550
256 512

we want 2926th term so 2926-2550=376
in 256 there will be 1's after there will be 2 so 2926th term will be 2

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