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06 Jan 2021, 06:50
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (02:15) correct
26%
(02:33)
wrong
based on 298
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of \(k\) if the sum of consecutive odd integers from 1 to \(k\) equals 441?
A. 47
B. 41
C. 37
D. 33
E. 29
M06-28
A. 47
B. 41
C. 37
D. 33
E. 29
M06-28
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06 Jan 2021, 07:29
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Bunuel
Sum of first n consecutive positive integers = \(1+3+5+7+9...=n^2\)
So \(n^2=441=21^2....n=21\).
The answer is not 21 as this means there are 21 terms in the sequence of consecutive odd integers.
\(21^{st}\) odd integer = \(2*21-1=42-1=41\)
B
General Discussion
13 Jan 2021, 07:00
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Bunuel
Official Solution:
What is the value of \(k\) if the sum of consecutive odd integers from 1 to \(k\) equals 441?
A. 47
B. 41
C. 37
D. 33
E. 29
Consecutive odd integers represent an evenly spaced set (aka arithmetic progression). Now, the sum of the terms in any evenly spaced set is the mean (average) multiplied by the number of terms, where the mean of the set is \(\frac{\text{first term} + \text{last term}}{2}\).
\(\text{average}=\frac{\text{first term} +\text{last term}}{2}=\frac{1+k}{2}\);
\(\text{# of terms}=\frac{k-1}{2}+1=\frac{k+1}{2}\) (# of terms in an evenly spaced set is \(\frac{\text{last term} - \text{first term}}{\text{common difference}}+1\))
\(\text{sum} = \frac{1+k}{2}*\frac{k+1}{2}=441\). Simplify: \((k+1)^2=4*441\), so \(k+1=2*21=42\) giving \(k=41\).
