Given: \(a_1=2\), \(a_2=-3\), \(a_3=5\) and \(a_4=-1\). Also \(a_n=a_{n-4}\), for \(n>4\).

Since \(a_n=a_{n-4}\) then:
\(a_5=a_1=2\);
\(a_6=a_2=-3\);
\(a_7=a_3=5\);
\(a_8=a_4=-1\);
\(a_9=a_5=a_1=2\);
and so on.

So we have groups of 4: {2, -3, 5, -1}, the sum of each of such group is \(2-3+5-1=3\). 97 terms consist of 24 full groups plus \(a_{97}=a_1=2\), so the sum of first 97 terms of the sequence is \(24*3+2=74\).

Answer: B.
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Given: a(n) = a(n-4) i.e. the nth term is same as (n-4)th term e.g. 5th term is same as 1st term. 6th term is same as 2nd term etc

The sequence becomes: 2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1 ...
The sum of each group of 4 terms = 2-3+5-1 = 3
How many such groups of 4 are there in the first 97 terms? When you divide by 4, you get 24 as quotient and 1 as remainder.
This means you get 24 complete groups of 4 terms each and 1 extra term i.e. the 97th term.
Sum of 24 groups of 4 terms each = 24*3 = 72
The 97th term will be 2 since it is the first term of the next group of four terms.
Sum of first 97 terms = 72+2 = 74
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can you explain a bit more.
from a5 to a93 i get an AP for which the sum is = 4371
from a1 to a4, the sum = 3
Total sum= 4374.
what am i doing wrong?

Nice one, a pattern problem with an addition twist.

a1 = 2, a2 = -3, a3 = 5, a4 = -1
What an = a(n-4) means that from a5 onwards, every number in the sequence will be equal to 4 numbers preceding that number. Therefore, a5 = a1 = 2, a6 = a2 = -3,...

Notice that this is a 4-element set pattern. If we sum a1-a4, we get 3. What is the closet multiple of 4 to 97 - 96! So, there are 24 such complete sets. Added to that is 97th number, which is first in the sequence = 2. So, we get (24 x 3) + 2 = 74 -> B
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From the equation, a5 = a1, a6 = a2, a7 = a3, a8 = a4 and again a9 =a5 =a1
So sum of 97 terms =sum of 96 terms +97th term
= 24*( a1 +a2 +a3 +a4) + a1
= 24*(2 -3 +5 -1) + 2
= 72 +2
= 74. Option B
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"\(a_n = a_{n-4}\) for n > 4." <--- what is the point of this information?
You could still solve for the problem without it, so what was the test makers intention on giving us it?

Anyway, Value of n = 1,2,3,4 | Value of An= (2) + (-3) + (5) + (-1) = 3; and for A5, A6, A7,A8, given n>4 = n-4, value for n again equals 1,2,3,4 | Value for An = 2) + (-3) + (5) + (-1) = 3
Therefore, the sum of the first 8 digits of 97 equals 6
97/8 = 12; 12 x 6 = 72
since the next digit in line is the start of a new cycle (i.e. equal to A1), that digits value is 2
thus 72 + 2 = 74 Answer (B)

But really, other than determining the next four digits, I do not know the purpose of the above piece of information; seeing as An-4 for the ninth digit would make A5, which no value was given, it was confusing at first - does the value of An, where n>=9, would repeat or not. I mean, giving the test structure, and cyclicality, I knew that I could stop at 8 and solve for it. However, it just seems weird since what is given really doesn't account for the value of A9-4 (A5) since they never give us the value of A5, you have to assume it repeats. It seems clearer to state this the question without An. But, I still solved it because I knew what they were looking for for this category of question.

12. Sequences

• Theory
  Math : Sequences & Progressions
  
  • Questions
    Special Numbers and Sequences Problems
    DS Questions
    PS Questions

For ALL other subjects check Ultimate GMAT Quantitative Megathread

(I pointed you to this topic multiple times)

Hope it helps.
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