What is the sum of the integers from 100 to 200 inclusive

Question

What is the sum of the integers from 100 to 200 inclusive which are divisible from 6 but not from 12?

1) 1200
2) 1350
3) 1850
4) 2550
5) 2650

PKN · Accepted Answer

What is the sum of the integers from 100 to 200 inclusive which are divisible   by 6 but not by 12  ?

If you observe the pattern of the numbers that are divisible by 6 but not by 12 in the given range, they lie at a distance of 12 from each other consecutively.

So the series of the terms starts at 102 and ends at 198, they form an AP with common difference of 12.

t_{n}=a+(n-1)*d=102+(n-1)*12 
Or, 198=102+(n-1)*12
Or, n=9

Sum,  S_{9}=\frac{n}{2}*(a+l) = \frac{9}{2}*(102+198)=\frac{9}{2}*300=9*150=1350

Ans. (B)

CounterSniper · Answer

We get the following numbers 
102, 114,126...

a common difference of 12

a1 =102
tn =200
200 = 102 + (n-1) 12
98 =12n -12
n = 110/12 = 9.6 
n = 9

sum = 9/2 ( 2*102 + (8)12) = 150 * 9 = 1350

JeffTargetTestPrep · Answer

This means we need to find the sum of the odd multiples of 6 from 100 to 200, inclusive, since even multiples of 6 will be divisible by 12. The first odd multiple of 6 within the range is 6 x 17 = 102, the next one is 6 x 19 = 114, etc. The last one is 6 x 33 = 198. So the odd multiples of 6 within the range are: 102, 114, 126, …, 198.

This is an arithmetic sequence of integers with a common difference of 12, so the number of integers is

(198 - 102)/12 + 1 = 96/12 + 1 = 9

The average of the integers is (198 + 102)/2 = 300/2 = 150. Thus, the sum is

150 x 9 = 1350

Answer: B/2

BrentGMATPrepNow · Answer

If you're unfamiliar with the formulas for arithmetic series, you can also answer the question this way...

We want the sum:  102  +  114  + 126 + . . . +  186  +  198  
If we add the terms in PAIRS (starting at each end), we get: ( 102  +  198  ) + ( 114   +  186  ) + . . . 
Simplify to get: 300 + 300 + ....

The question now is, "How many terms are in the sum 102 + 114 + 126 + . . . 186 + 198?

Notice that 102 = (17)(6) = (2 x  8  + 1)(6)
and 114 = (19)(6) = (2 x  9  + 1)(6)
and 126 = (21)(6) = (2 x  10  + 1)(6)
.
.
.
and 126 = (33)(6) = (2 x  16  + 1)(6)

We can see that the number of terms = the number of integers from  8  to  16  inclusive

A nice rule says:   the number of integers from x to y inclusive equals y - x + 1

So, the number of terms =  16  -  8  + 1 = 9
So, there are 9 terms, which means we have 4 pairs that add to 300, and 1 value, which we can deduce must be HALF of 300

So, the SUM = (4)(300) + 150 = 1350

Answer: B

Cheers,
Brent

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

What is the sum of the integers from 100 to 200 inclusive

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

What is the sum of the integers from 100 to 200 inclusive

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards