Problem Solving

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?

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.

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.

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.