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There are 50 odd terms and 49 even terms.

Arvind42
Bunuel
In a sequence \(a_1\),\(a_2\),…, each term after the first is found by taking the negative of the preceding term, and adding 1. If \(a_1 = 2\), what is the sum of the first 99 terms?

(A) 49
(B) 50
(C) 51
(D) 99
(E) 101

a1=2,a2=-1,a3=2,a4=-1....; So all odd values of n will be 2 and all even values will be -1; 49 Odd terms and 48 even terms; 49*2-48=50 IMO B
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Bunuel
In a sequence \(a_1\),\(a_2\),…, each term after the first is found by taking the negative of the preceding term, and adding 1. If \(a_1 = 2\), what is the sum of the first 99 terms?

(A) 49
(B) 50
(C) 51
(D) 99
(E) 101

from the given series we can say that odd no are 2 and even are -1
so total odd from 1 to 99 ; 50 and even ; 49
so 50*2+49*(-1) = 51
IMO C
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Bunuel
In a sequence \(a_1\),\(a_2\),…, each term after the first is found by taking the negative of the preceding term, and adding 1. If \(a_1 = 2\), what is the sum of the first 99 terms?

(A) 49
(B) 50
(C) 51
(D) 99
(E) 101

Let's list some terms to get a better understanding of the desired sum.
\(a_1 = 2\)
\(a_2 = -2 + 1 = -1\)
\(a_3 = 1 + 1 = 2\)
\(a_4 = -2 + 1 = -1\)
\(a_5 = 1 + 1 = 2\)
\(a_6 = -2 + 1 = -1\)
.
.
.
We can see that all ODD numbered terms are 2, and all EVEN numbered terms are -1

From 1 to 99 inclusive, there are 50 ODD numbered terms and 49 EVEN numbered terms

So, sum = (50)(2) + (49)(-1)
= 100 + (-49)
= 51

Answer: C

Cheers,
Brent
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Bunuel
In a sequence \(a_1\),\(a_2\),…, each term after the first is found by taking the negative of the preceding term, and adding 1. If \(a_1 = 2\), what is the sum of the first 99 terms?

(A) 49
(B) 50
(C) 51
(D) 99
(E) 101

We see that a1 = 2, a2 = -2 + 1 = -1, a3 = 1 + 1 = 2, a4 = -2 + 1 = -1 and so on.

Thus we can see that the odd-numbered terms are all 2 and the even-numbered terms are all -1. Since in the first 99 terms, there are 50 odd-numbered terms and 49 even-numbered terms, then the sum of the first 99 terms is:

50(2) + 49(-1) = 100 - 49 = 51

Answer: C
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