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31 Mar 2021, 02:30
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D
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Difficulty:
5%
(low)
Question Stats:
95% (01:21) correct
5%
(01:59)
wrong
based on 110
sessions
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If each number of a sequence is 4 more than the previous number, and the 3rd number in the sequence is 13, what is the 114th number in the sequence?
A. 117
B. 257
C. 444
D. 457
E. 477
A. 117
B. 257
C. 444
D. 457
E. 477
31 Mar 2021, 02:52
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1st term a = 5
114th term = a + (n - 1)d
= 5 + (114 - 1) 4
= 5 + 113 x 4
= 5 + 452
= 457
Option D
Posted from my mobile device
114th term = a + (n - 1)d
= 5 + (114 - 1) 4
= 5 + 113 x 4
= 5 + 452
= 457
Option D
Posted from my mobile device
15 Apr 2021, 11:26
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Bunuel
Solution:
We see that the sequence is an arithmetic sequence. In an arithmetic sequence, if d is the common difference, a is the nth term and b is the mth term, then
b = a + d(m - n)
Here, we can let b be the 114th term and we have:
b = 13 + 4(114 - 3)
b = 13 + 4(111)
b = 13 + 444
b = 457
Answer: D
