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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Difficult Math Problems #28 - Consecutive integers
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gamemaster
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800guy
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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
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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anandsebastin
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C800guy 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
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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BTGmoderatorRO
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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
(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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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.
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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Scott@TargetTestPrep
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We can label each integer as follows: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
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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