The sum S of the first n consecutive positive even integers is given

Question

The sum S of the first n consecutive positive even integers is given by S = n(n + 1). For what value of n is this sum equal to 110 ?

A. 10
B. 11
C. 12
D. 13
E. 14

(PS02978)

ravikumarmishra · Answer

10*11 = 110 so answer is A. No need to solve quadratic as options are very simple to test.

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Bismarck · Answer

OA:A

Given n(n + 1)=110, find n.
Plugging in the options 
first option,n=10
L.H.S= 10*11=110
R.H.S= 110
L.H.S=R.H.S
hence n =10

KSBGC · Answer

Formula = n(n+1)

as per the question,

n(n+1) = 110

Back solve it. Only option A matches .

The best answer is A.

Abhishek009 · Answer

n(n + 1) = 10*11

Thus, n = 10 Answer must be (A)

SSM700plus · Answer

Try plug-in.

A) 
if  n=10  , then  S = n(n+1) = 10 * 11 =110 .

Ans is A.

Kinshook · Answer

The sum S of the first n consecutive positive even integers is given by S = n(n + 1). For what value of n is this sum equal to 110 ?

S = n(n + 1) = 11*10 = 10*(10+1)
n=10

IMO A

KnockoutNed · Answer

Why do they use the term "consecutive positive even integers" if the integers in play are 10 and 11?

ManifestDreamMBA · Answer

Here 10 represents the number of terms n (i.e. there are 10 consecutive positive even integers being added)

n(n+1) is the formula for calculating the sum and hence 11 (n+1) comes into play

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