In a set of consecutive integers the least number is –23. If the sum

-23 + ( - 22) + ( - 21)............... 21 + 22 + 23 = 0

Now, after 23 there will be 2 numbers which will add upto 49 , ie 24 + 25 = 49

The greatest number is thus (C) 25

let n=number of integers in set
sum of n-1 integers=49+23=72
looking at answer choices,
24=72/3
let n-1=3
23+24+25=72
25
C

For Consecutive Integers Sum = ( last-first ) + 1
49 = L - (-23) + 1
49 = L + 23
L = 25
C

sum=N/2*(first term+ last term)

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Just curious to know if there is any other way to solve above question by using equations of Arithmetic progression

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This question is all about set that contains both positive and negative integers.

Problem i faced while solving this question is clueless set. amorphous set. Don't know anything about the set except sum and the least value.

Important thing is that if the set contains all the negative integers sum sum can't be positive. ****

It indicates that the set includes positive and negative integers.

now what about the no. of items.

set starts with -23. Thus , at least we have to have + 23 to overcome negative impact.

-23 +.......................................+ 23 =0

Given sum is 49.

we conclude that we have more elements in the set.

24 + 25 = 49.

Thus , 25 is the correct answer.

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