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1,2,2,4,4,4,8,8,8,8,......... find the 1000th term.

10 years ago

Notice that the no. of times the same no. are repeating forms an A.P.

Like, 1 is 1 times, 2 is 2 times, 4 is 3 times, 8 is 4 times and so on.

So, first we will find out which no. is repeated n times . This no. will be 1 term before that term which will appear in 1000th term.

So, the no. in the 1000th term will repeat (n+1) times.

As, sum of n natural no.'s is n(n+1)/2

Thus, n(n+1)/2 < 1000

or,   n^2 + n < 2000 < 2070(46*45)

Thus,  (n+46)(n-45)<0      Or,  n<45

So, n = 44

Thus, (n+1) = 45

Hence 1000th term will be included or appear in the no. that is written
(n+1) times or 45 times.

Notice that the series is 2^0, 2^1, 2^1,2^2,2^2,2^2 ,2^3,2^3,2^3,2^3 .....and so on.

So the no. that is repeated n times = 2^(n-1)

So 1000th term = 2^{(n+1)-1} = 2^n = 2^ 44