The sum of k consecutive integers is 41. If the least integer is -40

Question

The sum of k consecutive integers is 41. If the least integer is -40, then k =

A. 40
B. 41
C. 80
D. 81
E. 82

I have tried solving this problem by using formula sum of k consecutive integers = k(k+1)/2

Sum = 41
 But I am struggling to understand what role does the least integer play over here/. Can you please help?

Bunuel · Accepted Answer

The sum of k consecutive integers is 41 and the least integer is -40.

So, we have the following sum: (-40) + (-39) + ... = 41. The sum to be positive you must have positive numbers in the sequence as well. Each positive number and its mirror negative number will give the sum of 0, for example, -1+1=0, -2+2=0, -3+3=0, -40+40=0. Since the sum is 41, then the greatest number must be 41:

(-40) + (-39) + ... + (-1) + 0 + 1 + ... + 39 + 40  +  41  = 41 (the sum of red part is 0 + 41 = 41).

How many terms do we have? 40 negative numbers, 0 and 41 positive numbers, so total of 82.

Answer: E.

sanjoo · Answer

I think bunuel has explain it very nice !

-40........0.........41......total 82 integers.. -40 to 1-----40+1(0)+41(1 to 41)=82 ...

vijayshree · Answer

The sum of k consecutive integers is 41. If the least integer is -40, then k =

A. 40
B. 41
C. 80
D. 81
E. 82

Consecutive Integers ---The difference is consistent with each integers , therefore the series can be A.P .

Sum of A.P. = A + (N-1 ) D

A= First term 
D= Difference between each integer
N=number of terms

Sum = A + (N - 1 ) D
41= -40 + N -1 
82 = N

CONSIDER N=K

Answer = E

shravan2025 · Answer

Hello.
we can use direct formula = Sn =n/2(2a+(n-1)d
here sn =41, n =?( we have to find) a=-40 & d=1

after applying given value, we will get n=82 
so correct answer sis E

law258 · Answer

This is another good problem where it is helpful to visualize the number line:

We have -40 off in space on the left hand side of the line and we don't know how far out it goes on the right, but we do know that the sum is a positive number and its 41. Right away, you should think to yourself "okay, if I go out to 40 on the right hand side of the number line, everything on the left hand side will cancel it out." Therefore, if we add one more number, 41, on the right hand side we arrive at our value of 41 for the sum & the number of values on the line altogether is 41+40+1 =  82

rahul16singh28 · Answer

The Sum of an AP Series is  (n/2)  .. Given that  d = 1  &  Sum = 41 ,  a = -40 
Solving above we get,  n(n-81) = 82 .. No need to solve further.. n = 82

Abhishek009 · Answer

(-40 , -39 , -38.............-3 , -2 , - 1 , 0 , 1 , 2 , 3 ..............38 , 39 , 40 )  41

The result of the highlighted part will be 0 , and we have 81 Numbers in this range (-40 to 40 including 0)
 
So, The answer will be (E) 82

JeffTargetTestPrep · Answer

If the least (or first) integer is -40, we see that the sum of the consecutive integers from -40 to 40 will all cancel to zero, and if we include 41 as the final integer in the set, then the sum will be 41.

Thus, there are 41 - (-40) + 1 = 82 integers in total.

Answer: E

NTA · Answer

Let's have another approach,

41-(-40) = K-1
=> 81 = K-1
=> K= 82

Kinshook · Answer

Given: The sum of k consecutive integers is 41. 
Asked: If the least integer is -40, then k =

The sum of k consecutive integers = k/2 (2a + (k-1)) = k/2 (-80 + k - 1) = 41
k(k-81) = 82
k^2 - 81k - 82 = 0
(k-82)(k+1) = 0

k = 82

IMO E

Rance14 · Answer

I solved it using AP but it's much more convenient and faster to solve it by visualising and writing the numbers down

Using AP

Sn=n/2  -> sum of AP
41=k/2

you ultimately get a quadratic eqn -> k^2-81k-82=0

solving for k you get -> k=-1, k=82

But by writing down the numbers ->given sum=41

-40,-39,-38,-37,.......0,1,2,3......37,38,39,40,41

all the negatives and positives will cancel out except for 41
-40+40=0
-39+39=0
-38+38=0

ultimately leaving us with 0+41=41

so between -40 to 41 -> we have 82 numbers

Rance14 · Answer

I solved it using AP but it's much more convenient and faster to solve it by visualising and writing the numbers down

Using AP

Sn=n/2  -> sum of AP
41=k/2

you ultimately get a quadratic eqn -> k^2-81k-82=0

solving for k you get -> k=-1, k=82

But by writing down the numbers ->given sum=41

-40,-39,-38,-37,.......0,1,2,3......37,38,39,40,41

all the negatives and positives will cancel out except for 41
-40+40=0
-39+39=0
-38+38=0

ultimately leaving us with 0+41=41

so between -40 to 41 -> we have 82 numbers

bumpbot · Answer

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The sum of k consecutive integers is 41. If the least integer is -40

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The sum of k consecutive integers is 41. If the least integer is -40

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