Difficult Math Problems #28 - Consecutive integers

Question

OA coming after some people respond.. If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is: a) A+1 b) A+5 c) A+25 d) 2A e) 5A

gamemaster · Answer

n + n+1 + n+2 + n+3 + n+4 = 5n + 10 = A n+5 + n+6 + n+7 + n+8 + n+9 = 5n + 35 = A+25

800guy · Answer

here's the OA:

If you divide the sum obtained by adding any 5consecutive numbers by '5', and then you will get the Center number of the sequence itself.

i.e. 1 - 5 = 15/5 = 3 . 1, 2, 3, 4, 5

so, sixth consecutive number will be '3' more than the 'Middle term'
i.e. 3+3=6, similarly 3+4=7

Hence going by this. Asked sum would be

[(A/5) + 3]+[(A/5) + 4]+[(A/5) + 5]+[(A/5) + 6]+[(A/5) + 7] = A + 25

anandsebastin · Answer

800guy wrote:OA coming after some people respond..

If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a) A+1
b) A+5
c) A+25
d) 2A
e) 5A

C

Sum of 5 consecutive integers = x-2 +x-1+x+x+1+x+2 = 5x = A

Sum of next 5 consecutive integers = 5* (x+5)= 5x +25 = A+25

BTGmoderatorRO · Answer

for the sum of the first five, we have (n)(n+1)n+2(n+3)(n+4) = 5n+10 = A for the next five, we have (n+5)(n+6)(n+7)(n+8)(n+9) = 5n+35 = (5n + 10 ) + 25 = A+25 option C is the correct answer

Matt@VeritasPrep · Answer

It's easiest to see with numbers first.

If our first number is 1, then we start with

1, 2, 3, 4, 5

and the next six numbers are

6, 7, 8, 9, 10

Since each number is the second set is FIVE more than its counterpart in the first set (6 = 1 + 5, 7 = 2 + 5, 8 = 3 + 5, ...), we're adding 5 to our set FIVE times.

Adding 5 five times = adding 5 * 5 = adding 25.

Scott@TargetTestPrep · Answer

800guy wrote:
If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a) A+1
b) A+5
c) A+25
d) 2A
e) 5A

We can label each integer as follows:

First integer = x

Second integer = x + 1

Third integer = x + 2

Fourth integer = x + 3

Fifth integer = x + 4

Thus:

A = 5x + 10

A - 10 = 5x

(A - 10)/5 = x

A/5 - 2 = x

We see that the next 5 integers are x + 5, x + 6, x + 7, x + 8, and x + 9.

Since we have an evenly spaced set, sum = average x quantity, and since x = A/5 - 2, we have:

sum = (first term in the set + last term in the set)/2 x quantity

sum = [(A/5 - 2 + 5 + A/5 - 2 + 9)/2] x 5

sum = [(2A/5 + 10)/2] x 5

sum = (A/5 + 5) x 5

sum = A + 25

Alternate solution:

If we let the first integer be x, then the first five integers will be x, x + 1, x + 2, x + 3, and x + 4, and the next five integers will be x + 5, x + 6, x + 7, x + 8, and x + 9. We can see that each of the next five integers is 5 more than its counterpart in the first five integers. Thus, the sum of the next five integers is 5 x 5 = 25 more than the sum of the first five integers. So, if the sum of the first five integers is A, the sum of the next five integers is A + 25.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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Difficult Math Problems #28 - Consecutive integers

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