Math again?

Question

The least integer of a set of consecutives integers is -25. If the sum of this integers is 26, how many integers are in this set?

Thanks

Answer 1

Let the first integer = a
Number of terms = n
The terms are in the form a, a+1, a+2, a+3,....
The last term of this series is a+(n-1)
Sum of an AP series is
no of terms * (1st term + last term)/2
So n(2a+n-1)/2=Sum

now, a=-25,Sum =26
So n(-51+n)=52
or n^2 -51n-52=0
or (n-52)(n+1)=0
or n=52 or -1
but number of terms cannot be negative.
So n=52.
The numbers are -25,-24,-23,......0,1,2,....25,26

Answer 2

since it is a set of consecutive integers, the nos. in the set are in AP whose first term is -25 and sum is 26. let no. of nos. be 'n'. From the formula ...
Sum = n/2[2X(-25) + (n-1)d] where d is common difference which is 1,
thus, 26= (-50 + n - 1)n/2
or, 52 = -50n + n^2 -n
or, 52 = -51n +n^2
or, n^2 -51n - 52 =0
or, n^2 - 52n + n - 52=0
or, n(n-52) +1(n-52)=0
or, n=52 or n= 1
since n=1 is not possible
thus, n=52

Answer 3

The sum of -25 to 25 is zero. Just add 26.
-25 to 26 is 52 numbers

Answer 4

answer is 52,

ranging from -25.....0......+25....26

if u add all them integers up it equals 52