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30 Nov 2021, 05:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (01:41) correct
21%
(02:40)
wrong
based on 109
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Starting from 1, how many odd consecutive integers you need to sum in order to get 1600 as value?
A. 19
B. 20
C. 21
D. 40
E. 42
A. 19
B. 20
C. 21
D. 40
E. 42
30 Nov 2021, 06:36
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Formula for sum of sequence: \(S_n = \frac{n}{2}(a + l)\) where n is the number of terms, a is the first term and l is the last term.
First 5 odd numbers after:
1) 1
2) 3 = (2*1)+1
3) 5 = (2*2)+1
4) 7 = (2*3)+1
5) 9 = (2*4)+1
So every nth odd number can be written as 2(n-1)+1. Therefore l = 2(n-1)+1
\(1600 = \frac{n}{2}(1 + 2(n-1) + 1)\)
\(3200 = n(1+2n -2 +1) \)
\(3200 = 2n^2\)
\(1600 = n^2\)
\(n = 40\)
Answer D
First 5 odd numbers after:
1) 1
2) 3 = (2*1)+1
3) 5 = (2*2)+1
4) 7 = (2*3)+1
5) 9 = (2*4)+1
So every nth odd number can be written as 2(n-1)+1. Therefore l = 2(n-1)+1
\(1600 = \frac{n}{2}(1 + 2(n-1) + 1)\)
\(3200 = n(1+2n -2 +1) \)
\(3200 = 2n^2\)
\(1600 = n^2\)
\(n = 40\)
Answer D
09 Mar 2026, 13:38
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For this one, you just need to know an interesting fact: the sum of the first consecutive odd integers is simply k^2. Thus we have k^2 = 1600, k = 40, and thus our answer is 40! 
