Latest Exam Aptitude Question SOLUTION: find the sum of the series 1,2,1,2,2,1,2,2,2,1,2,2,2,2,.... till 1234th digit.

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(#M40040352) EXAM Numerical Ability question sum of the series Keep an EYE

find the sum of the series
1,2,1,2,2,1,2,2,2,1,2,2,2,2,.... till 1234th digit.

Asked In Exam harshvardhan singhch (9 years ago)

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Ram  (9 years ago)

1--2,1--2,2,1--2,2,2,1--2,2,2,2,1 and so on........
no. of terms=1,2,3,4,5------49=1225 terms+9 terms gives 1234 terms
so 1234th digit is 2
In 49 terms 1 will come 49 times=1*49=49
In 49 terms 2 will come 1176 times=2*[1+2+3+----48]=2352
(becz in 2nd terms 2 comes once and in 3rd terms 2 comes 2times and so on..
and adding we get 2+2*2+2*3+2*4+2*5--2*48=2352
In next 9 terms 2 only comes 9 times=2*9=18
so sum of series=49+2352+18=2419
Answer is 2419

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BHASKAR MANDAL  (9 years ago)

the position of occurance of 1 are
1,3,6,10,15,etc
so it is of the form n(n+1)/2
eg, the 10th position = 4(4+1)/2
here upto the 10th position the total no. of 1 occured = 4
ie, at position 1,3,6 and 10,
total = 4, it is the value of n when calculated by solving the above equation.
so the total no. of 1 that will be in the series is given by
n(n+1)/2 = 1234
on solving n = 50 or 49 , taking only integers.
when 50 is substituted in n
then n(n+1) /2 = 50*51/2= 2550
which is not possible
so the total no. of 1 in the series is 49
rest numbers will be 2
no. of 2 = 1234-49= 1185
total sum = 1185*2+49*1 = 2419
2419 is the answer

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