two consecutive numbers are removed from theprogression 1 2 3 n The arithmetic mean of the remaining number is 26

TCS Ans is 50
9 years agoHelpfull: Yes(3) No(0)
ans is 50
9 years agoHelpfull: Yes(1) No(0)
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..)
9 years agoHelpfull: Yes(1) No(4)
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
9 years agoHelpfull: Yes(1) No(0)
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.
9 years agoHelpfull: Yes(0) No(0)
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
9 years agoHelpfull: Yes(0) No(0)

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