Tricky Sequence and series question

[darden11]

The infinite sequence a1, a2 ....aN is such that a1=45, a2= 49 and a(n) = a(n-2) + 7 for n > 2. Which of the following is a term of the sequence?
Ø 323
Ø 463
Ø 498
Ø 533
Ø 598
This was a very tricky question that I came across. Would appreciate if somebody can solve it and help me understand that solution?

Disclaimer: -
I do not know the source of this question, so please do not ask it from me. A friend of mine who just wrote his GMAT and got a cool 750 gave me the pdf files containing some challenging questions. I have the answers to these questions but not the official explanations.

[dhoomketu]

We can write any of the elements of the series in the form of either:

a(x)=45+7(X) such as for a1, a3, a5 etc. or as;
a(x)=49+7(X) such as for a2, a4 etc;
here x goes from 0 to n

Now the thing is any number of the series from either 45 or 49 is subtracted should be divisible by 7. Let us look at each choice:

Ø 323--------Both (323-45) and (323-49) aren't divisible by 7
Ø 463--------Both not divisible by 7
Ø 498--------Both not divisible by 7
Ø 533--------Both not divisible by 7
Ø 598--------(598-45) is divisible by 7 so this could be the answer

Am I correct?

[Guest]

We can write any of the elements of the series in the form of either:

a(x)=45+7(X) such as for a1, a3, a5 etc. or as;
a(x)=49+7(X) such as for a2, a4 etc;
here x goes from 0 to n

Now the thing is any number of the series from either 45 or 49 is subtracted should be divisible by 7. Let us look at each choice:

Ø 323--------Both (323-45) and (323-49) aren't divisible by 7
Ø 463--------Both not divisible by 7
Ø 498--------Both not divisible by 7
Ø 533--------Both not divisible by 7
Ø 598--------(598-45) is divisible by 7 so this could be the answer

Am I correct?

Absolutely, my friend...u r answer is correct and thanks for the fantastic explanation. Would appreciate if u can post answers to the other remainder theory question that i had posted.

[AG]

The number in the series are of the following form

even numbers - div by 7

odd numbers - when divided by 7 remainder is 3

only E follows this patern.

[rfernandez]

Very good solution. Nice work.

In the future, per this forum's guidelines, please do not post questions unless you know their sources. Thank you for your understanding.

Rey