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02 Aug 2006, 17:34
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B
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Difficulty:
45%
(medium)
Question Stats:
70% (02:06) correct
30%
(02:25)
wrong
based on 1585
sessions
History
Date
Time
Result
Not Attempted Yet
The sum of the first \(n\) consecutive positive odd integers is equal to \(n^2\). What is the sum of all odd integers between 13 and 39, inclusive ?
A. 351
B. 364
C. 410
D. 424
E. 450
M20-31
A. 351
B. 364
C. 410
D. 424
E. 450
M20-31
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30 Sep 2020, 01:05
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Official Solution:
The sum of the first \(n\) consecutive positive odd integers is equal to \(n^2\). What is the sum of all odd integers between 13 and 39, inclusive ?
A. 351
B. 364
C. 410
D. 424
E. 450
The sum of all odd integers between 13 and 39, inclusive is equal to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive.
There are 20 odd integers from 1 to 39, inclusive, so the sum of all odd integers from 1 to 39, inclusive is \(20^2\).
There are 6 odd integers from 1 to 11, inclusive, so the sum of all odd integers from 1 to 11, inclusive is \(6^2\).
Thus, the required sum is \(20^2 - 6^2 = 364\).
Answer: B
The sum of the first \(n\) consecutive positive odd integers is equal to \(n^2\). What is the sum of all odd integers between 13 and 39, inclusive ?
A. 351
B. 364
C. 410
D. 424
E. 450
The sum of all odd integers between 13 and 39, inclusive is equal to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive.
There are 20 odd integers from 1 to 39, inclusive, so the sum of all odd integers from 1 to 39, inclusive is \(20^2\).
There are 6 odd integers from 1 to 11, inclusive, so the sum of all odd integers from 1 to 11, inclusive is \(6^2\).
Thus, the required sum is \(20^2 - 6^2 = 364\).
Answer: B
21 Jan 2015, 19:59
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Hi All,
These types of "sum of a sequence" questions can be approached in a number of different ways - you can use various formulas or you can approach this prompt by pattern-matching and minimizing the math that you have to do. For this question, you can use "bunching"....
We're dealing with a sequence of CONSECUTIVE ODD INTEGERS: 13 to 39, inclusive. We're asked for the SUM of this group.
1) Start with the sum of the smallest and the biggest: 13 + 39 = 52
2) Now look at the 'next smallest' and the 'next biggest': 15 + 37 = 52
From this, you can see that you're just going to be adding up a bunch of 52s. We DO have to check to see if there's a "middle" term in this sequence that doesn't get "bunched" though. To determine if that middle term exists, we just have find the last few 52s in the group....
21 and 31
23 and 29
25 and 27
Now we have proof that there is no middle term. We have 7 bunches of 52.
7(52) = 364
Final Answer:
GMAT assassins aren't born, they're made,
Rich
These types of "sum of a sequence" questions can be approached in a number of different ways - you can use various formulas or you can approach this prompt by pattern-matching and minimizing the math that you have to do. For this question, you can use "bunching"....
We're dealing with a sequence of CONSECUTIVE ODD INTEGERS: 13 to 39, inclusive. We're asked for the SUM of this group.
1) Start with the sum of the smallest and the biggest: 13 + 39 = 52
2) Now look at the 'next smallest' and the 'next biggest': 15 + 37 = 52
From this, you can see that you're just going to be adding up a bunch of 52s. We DO have to check to see if there's a "middle" term in this sequence that doesn't get "bunched" though. To determine if that middle term exists, we just have find the last few 52s in the group....
21 and 31
23 and 29
25 and 27
Now we have proof that there is no middle term. We have 7 bunches of 52.
7(52) = 364
Final Answer:
GMAT assassins aren't born, they're made,
Rich
