What is the sum of the multiples of 7 from 84 to 140

[Guest]

What is the sum of the multiples of 7 from 84 to 140, inclusive?

For this part of the solution: The mean of the set = (the first term + the last term) divided by 2: (84 + 140)/2 = 112
[would this work if the number of terms in the set was even?]

Thank you!

84 is the 12th multiple of 7. (12 x 7 = 84)
140 is the 20th multiple of 7.
The question is asking us to sum the 12th through the 20th multiples of 7.

The sum of a set = (the mean of the set) x (the number of terms in the set)
There are 9 terms in the set: 20th - 12th + 1 = 8 + 1 = 9
The mean of the set = (the first term + the last term) divided by 2: (84 + 140)/2 = 112 [would this work if the number of terms in the set was even?]
The sum of this set = 112 x 9 = 1008

Alternatively, one could list all nine terms in this set (84, 91, 98 ... 140) and add them.
When adding a number of terms, try to combine terms in a way that makes the addition easier
(i.e. 98 + 112 = 210, 119 + 91 = 210, etc).

The correct answer is C.

[unique]

What is the sum of the multiples of 7 from 84 to 140, inclusive?

For this part of the solution: The mean of the set = (the first term + the last term) divided by 2: (84 + 140)/2 = 112
[would this work if the number of terms in the set was even?]

Thank you!

84 is the 12th multiple of 7. (12 x 7 = 84)
140 is the 20th multiple of 7.
The question is asking us to sum the 12th through the 20th multiples of 7.

The sum of a set = (the mean of the set) x (the number of terms in the set)
There are 9 terms in the set: 20th - 12th + 1 = 8 + 1 = 9
The mean of the set = (the first term + the last term) divided by 2: (84 + 140)/2 = 112 [would this work if the number of terms in the set was even?]
The sum of this set = 112 x 9 = 1008

Alternatively, one could list all nine terms in this set (84, 91, 98 ... 140) and add them.
When adding a number of terms, try to combine terms in a way that makes the addition easier
(i.e. 98 + 112 = 210, 119 + 91 = 210, etc).

The correct answer is C.

I think this would work with even number of integers too. Because the mean is going to be x.5 (as you are averaging two integers).
That multiplied by the number of integers (which is even) will always result in an integer.

[StaceyKoprince]

Yep, you guys have got it!

[DCE]

Can you please exaplin this in detail?

Moreover, can you you also explain the logic behind:

Sum of a set = (the mean of the set) x (the number of terms in the set).

Thanks
DCE

[RonPurewal]

Can you please exaplin this in detail?

sure - what do you want explained in detail? (i.e., what is "this"?)

Moreover, can you you also explain the logic behind:

Sum of a set = (the mean of the set) x (the number of terms in the set).

this is just a rearrangement of the equation that DEFINES the mean of a set of numbers in the first place.
the definition of the mean is:
Mean of set = (sum of set) / (# terms in set)
if you multiply this equation by the # of terms in the set, you'll get the version above.