Latest TCS Aptitude Question SOLUTION: two consecutive numbers are removed from theprogression 1,2,3.......n. The arithmetic mean of the remaining number is 26(1/4). the value of n is ans.60

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two consecutive numbers are removed from theprogression 1,2,3.......n. The arithmetic mean of the remaining number is 26(1/4). the value of n is
ans.60

Asked In TCS harshit gupta (10 years ago)

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Sourav ganguly  (10 years ago)

mean= (n*(n+1)/2)/n = (n+1)/2
now (n+1)/2=26 1/4
solving for n,
n=50 apprx

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Kethineni Vinodh  (10 years ago)

TCS Ans is 50

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Theja Muppalla  (10 years ago)

ans is 50

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Neha Singh  (10 years ago)

after removing 2 nos from progression....n-2 elements will be left
sum of n-2 elements=((n-2)(n-1))/2
therefore arithmetic mean=((n-2)(n-1))/2(n-2)=26 1/4
n=53(approx..)

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nidhi sharma  (10 years ago)

As we can remove consecutive numbers from first end or last end only .because there is arithmetic mean of remaining term is given.if we remove from the middle progression will break.
1 case: when we remove 1 and 2 then arithmetic mean of(3,4,5...n) is 105 approx
2 case when we remove n and n-1 then (1,2....(n-1))
mean will be 53
hence 53 is d ans

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sanchari paul  (10 years ago)

Ans:50; The best method to answer the question is options method. Sum of first n natural numbers = n(n+1)/2. Sum of the numbers after removing 2 consecutive numbers = 105/4 * (n-2). Now n(n+1)/2 - 105/4 * (n-2) should be sum of two consecutive numbers. 50 satisfies this condition.

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Dhiren Paryani  (10 years ago)

if we remove 1,2 from the series it will look like this 3,4,....n

Using Sum of AP:

(n-2/2 X (3+n)) / n-2= 26 1/4

n=49.5

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