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

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

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

A) 896
B) 963
C) 1008
D) 1792
E) 2016

The OA is C.

Is there any way to solve it without listing the numbers?

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

A)896
B)963
c)1008
D)1792
E)2016

In other words, 84 + 91 + 98 + . . . 140 = ?

Since the values are EQUALLY SPACED, we can use the rule: SUM = [(FIRST + LAST)/2][# of values]

NUMBER of values
Here's a nice rule: If x and y are multiples of k, then the number of multiples of k from x to y inclusive = [(y-x)/k] + 1
So, for example, the NUMBER multiples of 3 from 6 to 21 inclusive = [(21 - 6)/3] + 1 = [15/3] + 1 = 6

So, the NUMBER multiples of 7 from 84 to 140 inclusive = [(140 - 84)/7] + 1
= [56/7] + 1
= 9

------------------------------------

Now apply the formula:
SUM = [(FIRST + LAST)/2][# of values]
= [(84 + 140)/2][9]
= [224/2][9]
= [112][9]
= 1008
= C

Cheers,
Brent

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We want 7*12 + 7*13 + ... * 7*20

To find this, we could add the first 20 multiples of 7, then subtract the ones we don't want (the first 11 multiples of 7).

That gives us:

(7*1 + ... + 7*20) - (7*1 + ... + 7*11)

From there, we could write this as:

7*(1 + 2 + ... + 20) - 7*(1 + 2 + ... + 11)

The two sums in parentheses are just triangular numbers, and the summation is a cinch. Smile

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

A) 896
B) 963
C) 1008
D) 1792
E) 2016

The OA is C.

Is there any way to solve it without listing the numbers?
Sum of the multiples of 7 from 84 to 140, inclusive = 84 + 91 + ... + 140.

Number of multiples of 7 from 84 to 140 = 140/7 - 84/7 + 1 = 20 - 12 + 1 = 9.

Now, write these numbers twice, once in ascending order and then in descending order:

S = 84 + 91 + .... + 133 + 140
S = 140 + 133 + .... + 91 + 84

The sum of each pair is 224. As the total number of pairs are 9.

2S = 224*9 or S = 112*9 = 1008.

Thus, the correct answer is C.

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What is the sum of the multiples of 7 from 84 to 140, inclusive?
Hi Vincen,
Let's take a look at your question.

First of all we need to find the number of terms between 84 to 140 inclusive.
Since we are talking about multiples of 7, lets find out the term numbers of 84 and 140 in the sequene of multiples of 7.
$$\frac{84}{7}=12$$
84 is 12th term.

$$\frac{140}{7}=20$$
140 is 20th term.

Total number of terms = n = 20 - 12 + 1 = 9

Sum of n terms in geometric series$$ =\frac{n}{2}\left(a_1+a_n\right)$$
$$=\frac{9}{2}\left(84+140\right)$$
$$=\frac{9}{2}\left(224\right)$$
$$=\frac{9}{2}\left(84+140\right)$$
$$=9\left(112\right)$$
$$=1008$$

Therefore, Option C is correct.
Hope this helps.

I am available, if you'd like any follow up.

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

A) 896
B) 963
C) 1008
D) 1792
E) 2016
We can use the following equation to determine the quantity:

(largest multiple in the set - smallest multiple in the set)/7 + 1 = number of multiples of 7

(140 - 84)/7 + 1

56/7 + 1 = 9

Next, let’s determine the average of this evenly spaced set.

average = (first number + last number)/2

average = (84 + 140)/2 = 224/2 = 112

Thus, the sum is 9 x 112 = 1,008.

Answer: C

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GMAT/MBA Expert

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

A) 896
B) 963
C) 1008
D) 1792
E) 2016

The OA is C.

Is there any way to solve it without listing the numbers?
In other words, 84 + 91 + 98 + . . . 140 = ?

Since the values are EQUALLY SPACED, we can use the rule: SUM = [(FIRST + LAST)/2][# of values]

NUMBER of values
Here's a nice rule: If x and y are multiples of k, then the number of multiples of k from x to y inclusive = [(y-x)/k] + 1
So, for example, the NUMBER multiples of 3 from 6 to 21 inclusive = [(21 - 6)/3] + 1 = [15/3] + 1 = 6

So, the NUMBER multiples of 7 from 84 to 140 inclusive = [(140 - 84)/7] + 1
= [56/7] + 1
= 9

------------------------------------

Now apply the formula:
SUM = [(FIRST + LAST)/2][# of values]
= [(84 + 140)/2][9]
= [224/2][9]
= [112][9]
= 1008
= C

Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
Use my video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for free Question of the Day emails
And check out all of these free resources

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