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Source: — Data Sufficiency |
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by nisagl750 » Thu Aug 30, 2012 3:15 am
shrey2287 wrote:122. An=An-2+11, n>2. Is 633 in the sequence?
1) A1=39
2) A2=43


OA : A
Since An=An-2+11, it represents an AP with d=11/2 = 5.5(A3=A1+11, A5=A3+11 and so on, So A2 = A1+5.5 & A3=A2+5.5),
nth term of an A.P sequence is given by N=a+(n-1)d, here N = 633 (we want to find whether is in the series or not)
Statement1: a=39, N=633, d=5.5
633=39(n-1)5.5
5.5(n-1)=594
(n-1)=108
n=109
So 633 is 109th term of the series if A1=39 Sufficient

Statement2: 43=39, N=633, d=5.5
633=43(n-1)5.5
5.5(n-1)=590
(n-1)=107.272727.......
n=108.2727.....
since n is a number, it can't be in decimal. Insufficient

Hence, A
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by neelgandham » Thu Aug 30, 2012 3:27 am
An=An-2+11, n>2. Is 633 in the sequence?
1) A1=39
Let n = 3 then A3 = A1 + 11. A3 = 39 + 11
A5 = A3 + 11 = 39 + 11 + 11 = 39 + 2*(11)
A7 = A5 + 11 = 39 + 2*(11) + 11 = 39 + 3*(11)
Now, we know that An = 39 + [(n-1)/2]*11, where n is the number of the term and a positive integer.
Let us check if 633 is one of the terms
633 = 39 + [(n-1)/2]*11
594 = [(n-1)/2]*11
54 = [(n-1)/2]
n = 109. So there exists an n which is an integer and 633 is a term in the series. So, statement I is sufficient to answer the question.
post statement - We don't know the values of the even terms.
2) A2=43
An=An-2+11
A4 = A2 + 11 = 43 + 11
A6 = A4 + 11 = 43 + 11 + 11 = 43 + 2*11
A8 = A6 + 11 = 43 + 2*11 + 11 = 43 + 3*11
So, An = 43 + [(n-2)/2]*11
Let us check if 633 is one of the terms
633 = 43 + [(n-2)/2]*11
590 = [(n-2)/2]*11
(1180/11) + 2 = n = Not an integer.
So, 633 is not one of the EVEN terms. Since we are not sure if 633 is one of the ODD terms, statement II is insufficient to answer the question.

IMO A
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by Anurag@Gurome » Thu Aug 30, 2012 3:40 am
shrey2287 wrote:
122. An=An-2+11, n>2. Is 633 in the sequence?
1) A1=39
2) A2=43


OA : A
An = An-2 + 11, n > 2
A3 = A1 + 11, A5 = A3 + 11.... (so the odd numbers are in AP with initial term as A1 and common difference as 11)
A4 = A2 + 11, A6 = A4 + 11 ... (so the even numbers are also in AP with initial term as A2 and common difference as 11)
But the whole series taken is not in AP with common difference 5.5.
The series can be thought as two AP series with terms one term from one AP and the next term from another AP Series. but both has common difference 11.
The series can be written as a,b,a+11,b+11,a+22,b+22 (a1,b1,a2,b2,a3,b3)...

consider 1st option,
it is given a as 39
633 = 39 + (n-1)11
n = 54
so it will be a54, which is A107 of the original series.
if it is a1,a2,a3,.. series then it will be in A1,A2,A3... series too.

consider 2nd option,
633 = 43 + (n-1)11
n = 53.63 which is not an integer
so it is not part of b1,b2,b3.. series and we can`t answer whether it is a1,a2,a3.. series

Hence, It is A
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by nisagl750 » Thu Aug 30, 2012 3:49 am
Anurag@Gurome wrote: But the whole series taken is not in AP with common difference 5.5.
Got it, thanks
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