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Difficult Math Problems #28 - Consecutive integers

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by 800guy » Sun Oct 15, 2006 1:01 pm
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
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by gamemaster » Mon Oct 16, 2006 6:34 am
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
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OA

by 800guy » Tue Oct 17, 2006 4:40 pm
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
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Re: Difficult Math Problems #28 - Consecutive integers

by anandsebastin » Wed Oct 18, 2006 6:27 pm
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
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by BTGmoderatorRO » Sat Sep 16, 2017 7:00 pm
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
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by Matt@VeritasPrep » Sun Sep 17, 2017 5:14 pm
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.
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by Scott@TargetTestPrep » Tue Dec 12, 2017 4:12 pm
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

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