if by beginning with one consecutive numbers are continuously return to its right then which digit will be written on 31st position
Answers
Answer:
After one value is removed:
Since all of the values are Integers, the sum here must be an integer.
Sum=(number)×(average).
Since the average =35
17
7
, and the sum must be an integer, the number of integers must be a multiple of 17.
For any evenly spaced set, average = median.
After one of the consecutive integers is removed, most of the remaining set will still be evenly spaced.
As a result, the average of the remaining set 37
17
7
will still be close to the median.
Implication:
The number of integers =4×17=68, with the result that 35
17
7
will be close to the median of the 68 mostly consecutive integers.
∴ Sum=68×35
17
7
=2408.
Original set:
Since 68 integers remain after one of the integers is removed, the original set contains 69 integers.
Sum of the first n positive integers =
2
(n)(n+1)
.
∴ Sum=69×
2
70
=2415.
Removed integer = original sum - sum after one integer is removed
=2415−2408=7.
Step-by-step explanation:
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Answer:
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