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07 Aug 2024, 21:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
68% (02:13) correct
32%
(02:38)
wrong
based on 234
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Let \(a_1, a_2, a_3, a_4, a_5\) be a sequence of five consecutive odd integers. Consider a new sequence of five consecutive even integers ending with \(2a_3\). If the sum of the numbers in the new sequence is 450, then \(a_5\) is
A. 43
B. 45
C. 47
D. 49
E. 51
A. 43
B. 45
C. 47
D. 49
E. 51
08 Aug 2024, 01:30
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Assuming the initial sequence to be a, a+2, a+4, a+6 and a+8.
In this case a3 will be a+4 and thus 2a3 will be 2a+8.
In the new sequence, 2a+8 should be the last of 5, so the new sequence is 2a, 2a+2, 2a+4 2a +6 and 2a+8.
The summation of the above is 450 so 10a + 20 is 450 and a is 43.
Now we need a5 which is basically a+8 which is 43+ 8 = 51
Answer - Option "E"
In this case a3 will be a+4 and thus 2a3 will be 2a+8.
In the new sequence, 2a+8 should be the last of 5, so the new sequence is 2a, 2a+2, 2a+4 2a +6 and 2a+8.
The summation of the above is 450 so 10a + 20 is 450 and a is 43.
Now we need a5 which is basically a+8 which is 43+ 8 = 51
Answer - Option "E"
15 Sep 2024, 12:34
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we know that
ev1 + ev2 + ev3 + ev4 + ev5 = 450
knowing that they are all consecutive even numbers
ev2 = ev1 +2
ev3 = ev1 + 4
and so on...
so we have
5ev1 + 20 = 450 --> ev1 = 86
now we know that ev5 = 2*a3 (where a3 represent the third entry in the consecutive ODD sequence)
so a3 = 94/2 = 47
now if a3 is 47 and it is the third entry in a sequence of CONSECUTIVE odd numbers, then a4 and a5 will be:
a4 = 49
a5 = 51
ev1 + ev2 + ev3 + ev4 + ev5 = 450
knowing that they are all consecutive even numbers
ev2 = ev1 +2
ev3 = ev1 + 4
and so on...
so we have
5ev1 + 20 = 450 --> ev1 = 86
now we know that ev5 = 2*a3 (where a3 represent the third entry in the consecutive ODD sequence)
so a3 = 94/2 = 47
now if a3 is 47 and it is the third entry in a sequence of CONSECUTIVE odd numbers, then a4 and a5 will be:
a4 = 49
a5 = 51
