Question

a series 1234123441234441234444.....n what is the value on 200th position..??

Answer 1

Given series is 1234123441234441234444.... n
we can write it ,
$\underbrace{1234}\underbrace{12344}\underbrace{123444}\underbrace{1234444}.......n$

we see , 1234 => 4 terms
12344 => 5 terms
123444 => 6 terms
1234444 => 7 terms and so on...

we have condition is {4 terms } + {5 terms } + {6 terms } + {7 terms } + ...... < 200terms
so, 4 + 5 + 6 + 7 + ..... n < 200
(1 + 2 + 3 )+ 4 + 5 + 6 + 7 + .... n < 200 + 1 + 2 + 3
n(n + 1)/2 < 200 + 6 = 206
n(n + 1) < 412 it is possible when n ≤ 19 [ because n is integers so I assume only integers value of n that's why n ≤ 19]
so, n = 19
so, Sn = 4 + 5 + 6 + 7 + ... + 19
Sn = 19(19 + 1)/2 - (1 + 2 + 3) = 184

hence, after 184 terms , sequence has 20 terms .e.g., 12344444444444444444
hence, 200th term is not other than 4

hence, 4 is in 200th position.

Answer 2

1234 => 4 times

12344 => 5 times

123444 => 6 times

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4 + 5 + 6 + ...n < 200

n(n+1)/2 - 6 = 200

So n = 19

i.e. 4+5+6+...19 = 184

After the 184th term,the sequence has 20 terms

12344444444444444444 = 20 terms

=> 200th term = 4