Question

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​

Answer 1

Answer:

Given: a1=2a1=2, a2=−3a2=−3, a3=5a3=5 and a4=−1a4=−1. Also an=an−4an=an−4, for n>4n>4.

Since an=an−4an=an−4 then:

a5=a1=2a5=a1=2;

a6=a2=−3a6=a2=−3;

a7=a3=5a7=a3=5;

a8=a4=−1a8=a4=−1;

a9=a5=a1=2a9=a5=a1=2;

and so on.

So we have groups of 4: {2, -3, 5, -1}, the sum of each of such group is 2−3+5−1=32−3+5−1=3. 97 terms consist of 24 full groups plus a97=a1=2a97=a1=2, so the sum of first 97 terms of the sequence is 24∗3+2=7424∗3+2=74.

Answer: B.

Answer 2

Answer:

A. 72

Step-by-step explanation:

see in pic mate