What is the 999th term of the series S ? (1) The first 4

Question

4. What is the 999th term of the series S ?

(1) The first 4 four terms of S are (1 + 1)Â² , (2 + 1)Â² , (3 + 1)Â² , and (4 + 1)Â².
(2) For every x, the xth term of S is (x + 1)Â².

FN · Answer

D? or is there a catch?

ninomoi · Answer

I think it is D.

Either statement can answer the question.

If the 1st term => (1+1)^2 and 2nd term (2+1)^2

Then the sequence should be like this:

For nth term=> (n+1)^2

This also answers stmt 2

Hope this helps

techjanson · Answer

I think there is a catch.
(1) is SUFF
but in (2) it says that for every x xth term is (x+1)^2 so when x is 2, 2nd term is (2+1)^2, for 3, 3rd term is (3+1)^2 and so only for 999 we know the 999th term which seems INSUFF

The wording in (2) seems weird.

ninomoi · Answer

You know what you maybe right! ANy other explanations?!

ricokevin · Answer

B?

(1) The first 4 four terms of S are (1 + 1)Â² , (2 + 1)Â² , (3 + 1)Â² , and (4 + 1)Â².
INSUFFICIENT. You only know the first four terms. Unless there's a formula such as f(n)=n+2 (made up randomly) that applies to all numbers, you cannot assume that terms after the first 4 follow what apparently seems like nth term = (n+1)^2.

(2) For every x, the xth term of S is (x + 1)Â².
SUFFICIENT. It's worded weirdly but I think "for every x" means "for any number (integer probably) x that refers to the term in S."

Hmmm...

Fig · Answer

(B) as well

Same reasoning

.... U could have a sequence that is adapted to a diferrent rule for any term in it

What is the 999th term of the series S ? (1) The first 4

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