Member of the Sequence

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Member of the Sequence

by rahulvsd » Sun Oct 23, 2011 8:45 am
A sequence of numbers is such that a1 = 11, a2 =16, and each subsequent an = an-2 + 9. Which of the following numbers is a member of the sequence?

A. 216
B. 246
C. 289
D. 299
E. 368

OA: D
Source: — Problem Solving |

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by knight247 » Sun Oct 23, 2011 10:44 am
From the formula given we can deduce that

a1=11
a2=16
a3=20
a4=25

etc
Lets separate the odd numbered terms from the even numbered ones

a1=11 a2=16
a3=20 a4=25
a5=29 a6=34

etc

For the odd numbered terms, starting from 11, the number 9 is added to every subsequent term. And for the even numbered terms, starting from 16 the number 9 is added to every subsequent terms.

Now, whether any term An of the sequence is an odd or even numbered term, the number of terms from A1 or A2 (Depending on whether An is an ODD or EVEN term) to An is going to be an Integer..Obviously!! Which of the answer choices meets that requirement??

A. For Odd numbered terms [(216-11)/9]+1 NO INTEGER VALUE
For even numbered terms [(216-16)/9]+1 NO INTEGER VALUE

B. For Odd numbered terms [(246-11)/9]+1 NO INTEGER VALUE
For even numbered terms [(246-16)/9]+1 NO INTEGER VALUE

C. For Odd numbered terms [(289-11)/9]+1 NO INTEGER VALUE
For even numbered terms [(289-16)/9]+1 NO INTEGER VALUE

D. For Odd numbered terms [(299-11)/9]+1 YESS!!!! INTEGER VALUE!! This means that 299 is one of the odd numbered terms.
For even numbered terms [(299-16)/9]+1 NO INTEGER VALUE

E. For Odd numbered terms [(368-11)/9]+1 NO INTEGER VALUE
For even numbered terms [(368-16)/9]+1 NO INTEGER VALUE

Hence, D

BTW, the formula I've used to determine the number of NUMBERS is as follows,

Number of NUMBERS with a constant increment within a certain range=
[(Biggest Number-Smallest Number)/Increment]+1

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by neelgandham » Sun Oct 23, 2011 1:06 pm
From the question

a1 = 11
__a2 = 16
a3 = 20
__a4 = 25
a5 = 29
__a6 = 34

From the above we can tell that the numbers in the sequence are of the form 11+9*n or 16+9*m, where m,n>=0 and m,n are integers.

A. 216
11+9n = 216 => 9n = 205, n is not an integer
16+9m = 216 => 9m = 200, m is not an integer

B. 246
11+9n = 246 => 9n = 245, n is not an integer
16+9m = 246 => 9m = 240, m is not an integer

C. 289
11+9n = 289 => 9n = 278, n is not an integer
16+9m = 289 => 9m = 273, m is not an integer

D. 299
11+9n = 299 => 9n = 288, n is an integer, and n = 32
16+9m = 299, Need not verify !

E. 368
Forget it :)

Answer : D
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by VivianKerr » Sun Oct 23, 2011 10:12 pm
The key is to look for a pattern: 11, 16, 20, 25....

Notice how as we go, we are basically just adding 9's to 11, and adding 9's to 16.

So any number in the sequence will either be equal to 11 + 9x, or 16 + 9x.

We can start by dividing 9 into each answer choice and checking the remainders, then seeing if we can get either an 11 or a 16 by adding one more 9 to the remainder.

216/9 = 24, no remainder (so no possibility for a 11 or 16)

246/9 = 27, remainder 3. Even if we add 9 to that, we get 12 (so no poss. 11 or 16)

289/9 = 32, remainder 1. Adding a 9 only gives us 10.

299/9 = 33, remainder 2. Adding 9 to that gives us 11!! We have found a possible number.
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