multiples of 7

This question tests your knoweldge of AP series

84 = 12x7 .. so 84 is 12the term
140= 20 X7 ..... 140 is 20th term ....

so total number of terms between 84 and 140 = 9 . ( using the formula total numbers between b and a inclusisve is( b-a ) +1

now the sum of an AP series is... n/2 ( first term + last Term ) = 9/2 (84 +140 ) = 1008

Answer C .

Brent@GMATPrepNow wrote:

j_shreyans 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

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)/3] + 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

Hi Brent,

I think here
So, the NUMBER multiples of 7 from 84 to 140 inclusive = [(140 - 84)/3] + 1

you mean 7 in lieu of 3?

You're absolutely right. Good catch???


waleeed

When it comes to calculating the number of item in a sequence of multiples, i find it easy to divide first term and last term by the multiple then apply the formula of (last-First+1).

In this case:
84/7=12
140/7=20
#items=20-12+1=9
from here you can follow any approach that makes you comfortable.

j_shreyans 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

OAC

How to approach this kind of question?

Solution:

Sum of the multiples of 7 from 84 to 140, inclusive,
84 + 91 + 98 + 105 + ... +140

This above series represents an arithmetic series with a common difference d = 7.
We know that sum of a finite arithmetic series can be calculated using the formula,

Sum = (First term + Last term)(n/2) where n is the number of terms in the series.
Therefore, to find sum, we need to find the number of terms in the series above.

We know that the terms in the arithmetic sequence are the multiples of 7. Therefore,
First term = 84 = 7 x 12 = 12th multiple of 7
Last term = 140 = 7 x 20 = 20th multiple of 7

The number of terms in the series 84 + 91 + 98 + 105 + ... +140
= 9 (84 and 140 inclusive)
Therefore, n = 9

Now,
Sum = (First term + Last term)(n/2)
Sum = (84 + 140)(9/2)
Sum = (224)(9/2)
Sum = (224 x 9) / 2
Sum = 2016 / 2
Sum = 1008

Therefore, Option C is the correct answer.

A streamlined solution:
84 / 7 = 12
140 / 7 = 20
So, it would be seven multiple of 12, 13, 14, 15, 16, 17, 18, 19 and 20.
Thus the sum of the multiples of 7 from 84 to 140, inclusive, is:
7(12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20) = 7 (144) = 1008.

Answer: C