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by Anurag@Gurome » Mon Dec 19, 2011 1:31 am
karthikpandian19 wrote:If a sequence of 8 consecutive odd integers with increasing values has 9 as
its 7th term, what is the sum of the terms of the sequence?
(A) 22 (B) 32 (C) 36 (D) 40 (E) 44
Let the 1st term of the sequence = a
Since the sequence is of consecutive odd integers, so the common difference, d = 2, which implies that the sequence will be an arithmetic series.
Hence, Sum of the terms = (n/2)[2a + (n - 1)d]
Now, a(7) = 9 implies a + 6d = 9
a + (6 * 2) = 9, as d = 2
a = -3
So, required sum of 8 terms = (8/2)[-6 + (7 * 2)] = 4 * 8 = 32

The correct answer is B.
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by GMATGuruNY » Mon Dec 19, 2011 3:54 am
karthikpandian19 wrote:If a sequence of 8 consecutive odd
integers with increasing values has 9 as
its 7th term, what is the sum of the
terms of the sequence?
(A) 22 (B) 32 (C) 36
(D) 40 (E) 44

Counting down from the 7th term: 9,7,5,3...
The median is the average of the 5th and 4th terms:
(5+3)/2 = 4.
Given evenly spaced values, sum = number*median:
8*4 = 32.

The correct answer is B.
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by sam2304 » Mon Dec 19, 2011 5:59 am
Sum = average * n

average = (first term + last term)/2

average = (11 - 3)/2 = 4

sum = 4 * 8 = 32

IMO B.
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by sk8legend408 » Mon Dec 19, 2011 6:09 am
You can also just draw it out.

Count backwards starting from the 7th term (9), 9+7+5+3+1+(-1)+(-3)=21.

You have to remember that the 8th term will be the next odd number after 9 which is 11.

21+11=32.
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by karthikpandian19 » Wed Dec 28, 2011 9:58 pm
OA is B
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by Abhishek009 » Thu Dec 29, 2011 7:43 am
karthikpandian19 wrote:If a sequence of 8 consecutive odd
integers with increasing values has 9 as
its 7th term, what is the sum of the
terms of the sequence?
(A) 22 (B) 32 (C) 36
(D) 40 (E) 44
Best way which I can think of is using a table as follows :


Image

Then add up the numbers to get the answer within 30 seconds...
Abhishek
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