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

My approach:
According to the formula:
a5 = an-4 = a1=2. Meaning that the cycle repeats after every four number.
And the sum of a cycle =2-3+5-1=3
So, 97/4 = 24 full cycle.
So sum will be 24*3 =72+the first term=72+2=74.

According to me, the answer is 74 (B). This is one way you could solve it -

Since the given sequence is

2, -3, 5 , 1,

From the formula given

a5 = a(5-4)=a1=2

Similarly a6=a2 (-3) , a7=a3 (5), a8=a4 (-1)

So the sequence of first four numbers is repeating -
Hence the infinite sequence looks like this:

2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1 and so on...

Now to find out the sum, you can group the pattern like this

(2, -3, 5 , -1) + (2, -3, 5, -1) + ...

Find sum of each group -
(3) + (3) + ....


Since the pattern is repeating after every four numbers, and since 96 is multiple of 4, the new pattern will start from 97th number.

97th number is 2

97 = 4 * n + 1

n=24



So the sum of 97 terms is

(3) * 24 + 2 = 72 + 2

Let me know if you have quest!

at least make the effort to post the question correctly

what is an = an-4

is it an = (an-4)
or is it an = a(n-4)