Sequence of numbers

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Sequence of numbers

by oquiella » Sat Oct 31, 2015 1:07 pm
96. If a sequence of numbers is given by An = An-1+15, is A32 an odd number?
(1) A1 is an odd integer
(2) A21 is an odd integer

How would you approach this problem?


Explanation:
On observing the sequence, it is clear that it is an arithmetic progression with common difference 15. So terms will be even and odd alternately.
Statement 1 says A1 is odd, and odd +15(odd) is even, so A2 will be even and A3 odd and so on. So A32 will be even; SUFFICIENT.
Similarly, from statement 2, it is clear that A21, A23, A25 and so on will be odd while A22, A24 and so on will be even. Hence, A32 will be even; SUFFICIENT.
The correct answer is D;
each statement alone is sufficient.
Source: — Data Sufficiency |

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by GMATGuruNY » Sun Nov 01, 2015 4:12 am
oquiella wrote:96. If a sequence of numbers is given by A(n) = A(n-1)+15, is A₃₂ an odd number?
(1) A� is an odd integer
(2) A₂� is an odd integer
The formula in red implies that each term after the first is 15 greater than the preceding term.
WRITE IT OUT and LOOK FOR A PATTERN.

Statement 1:
Let A� = 1.
Then:
A₂ = A� + 15 = 1+15 = 16.
A₃ = A₂ + 15 = 16+15 = 31.
A₄ = A₃ + 15 = 31+15 = 46.
The results in red indicate that every EVEN-NUMBERED TERM will be even.
Thus, A₃₂ is NOT odd.
SUFFICIENT.

Statement 2:
Let A₂� = 1.
Then:
A₂₂ = A₂� + 15 = 1+15 = 16.
A₂₃ = A₂₂ + 15 = 16+15 = 41.
A₂₄ = A₂₃ + 15 = 21+15 = 46.
The results in red indicate that every EVEN-NUMBERED TERM will be even.
Thus, A₃₂ is NOT odd.
SUFFICIENT.

The correct answer is D.
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