Akay, ur qs -

The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

So a5 = 5 - 4 = 1
a6 = 6 - 4 = 2
:
:
a94 = 94 - 4 = 90
So sum of this numbers = n/2(a + b) = 90 / 2 * (1 + 90) = 4095
where n = No of terms
a = First term
b = Last term
And sum of the first few terms; a1 , a2, a3 and a4 = 3
So total summation is = 4095 + 3 = 4098

But this is not provided in the options given.

I think the answer is B

4 terms are repeating.
Sum of the first 4 terms is 3 = 2-3+5-1

If you divide 97 by 4 = 24 + 1
Therefore, the sum of the first 97 terms is 3*24 + 2 = 74

97th term is same as first term which is 2 since 4 terms are repeating.

I am a little confused

As per the question an= an-4. Is this a(n-4) or (an)-(4)?

kajcha as per the question it is a(n-4) /* n-4 in subscript */

That's what I was thinking.. Ans should be B

Same explanation as josephcho77

Agreed.If an=a(n-4) meaning if n=5, a5=a(5-4)=a1=2 then a1+a2+a3+a4=3 meaning the addition of these 4 terms happens 97/4=24+1 (the "1" meaning a1).so ((24*3)+2)=74.Hence B

Sorry guys! I misunderstood the qs and took a wrong approach to solve it. Sorry one again for my misunderstanding!!!