Data Sufficiency

122. An=An-2+11, n>2. Is 633 in the sequence?
1) A1=39
2) A2=43


OA : A

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

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

Anurag@Gurome wrote: But the whole series taken is not in AP with common difference 5.5.

Got it, thanks