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Sub 505 Level|   Arithmetic|                  
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Bunuel
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Bunuel
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


NEW question from GMAT® Quantitative Review 2019


(PS02978)


Formula = n(n+1)

as per the question,

n(n+1) = 110

Back solve it. Only option A matches .

The best answer is A.
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Bunuel
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


NEW question from GMAT® Quantitative Review 2019


(PS02978)
\(n(n + 1) = 10*11\)

Thus, n = 10 Answer must be (A)
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Bunuel
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


NEW question from GMAT® Quantitative Review 2019


(PS02978)

Try plug-in.

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

Ans is A.
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Bunuel
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


NEW question from GMAT® Quantitative Review 2019


(PS02978)

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