LUANDATO wrote:In a sequence a1,a2,a3,..., each term after the first is found by taking the negative of the preceding term, and adding 1. If a1 = 2, what is the sum of the first 99 terms?

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

The OA is C.

I don't have very clear this PS question. Experts Can you help me to solve it please?

First find the pattern:
A1 = 2
A2 = -2 + 1 = -1
A3 = -(-1) + 1 = 2
A4 = -1

So our sequence looks like this: 2, -1, 2, -1, 2...
All the ODD terms = 2. # of ODD terms from A1 - A99 = 50. So 50 terms equal to 2 will give us 2*50 = 100.
All the EVEN terms = -1. # EVEN terms from A2 - A98 = 49. 50 terms equal to -1 = -49.
Total Sum = 100 + (-49) = 51.

The answer is C

In a sequence a1,a2,a3,..., each term after the first is found by taking the negative of the preceding term, and adding 1. If a1 = 2, what is the sum of the first 99 terms?

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

The OA is C.

I don't have very clear this PS question. Experts Can you help me to solve it please?

Hi LUANDATO,
Let's take a look at your question.

$$a_1=2$$
$$a_2=-2+1=-1$$
$$a_3=-\left(-1\right)+1=1+1=2$$
$$a_4=-2+1=-1$$
$$a_5=-\left(-1\right)+1=1+1=2$$

Since, all odd terms are 2 and all even terms are -1, therefore,
$$a_{99}=2$$

The sequence will be like,
$$2,\ -1,\ 2,\ -1,\ 2,\ -1,\ ...,\ 2$$
The sequence has 99 terms in total.
It has 49 pairs of two terms 2 and -1 and then a last term 2.
We can write the sum as,
$$Sum = \left(2-1\right)+\left(2-1\right)+\left(2-1\right)+....+\left(2-1\right)+2$$
$$Sum=\left(1\right)+\left(1\right)+\left(1\right)+....+\left(1\right)+2$$
Since, there are 49 ones in the sum above, therefore,
$$Sum=49+2$$
$$Sum=51$$

Therefore, Option C is correct.

Hope this helps.
I am available if you'd like any follow up.