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# In a sequence a1, a2, …, each term after the first is found by taking

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Joined: 02 Sep 2009
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In a sequence a1, a2, …, each term after the first is found by taking  [#permalink]

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12 Jun 2019, 02:06
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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

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In a sequence a1, a2, …, each term after the first is found by taking  [#permalink]

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12 Jun 2019, 02:29
First term=$$a_1$$
Second term = $$-a_1 + 1$$
Third term = $$a_1-1+$$1=$$a_1$$
fourth term = $$-a_1+$$1
And so on

the sum of the first 99 terms= $${49*[a_1+(-a_1+1)]}+a_1$$= 49+2=51

Bunuel wrote:
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
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In a sequence a1, a2, …, each term after the first is found by taking  [#permalink]

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Updated on: 12 Jun 2019, 21:41
Bunuel wrote:
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; 50 Odd terms and 49 even terms; 50*2-49=51 IMO B

Originally posted by Arvind42 on 12 Jun 2019, 02:35.
Last edited by Arvind42 on 12 Jun 2019, 21:41, edited 1 time in total.
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Re: In a sequence a1, a2, …, each term after the first is found by taking  [#permalink]

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12 Jun 2019, 02:37
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1
There are 50 odd terms and 49 even terms.

Arvind42 wrote:
Bunuel wrote:
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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Re: In a sequence a1, a2, …, each term after the first is found by taking  [#permalink]

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12 Jun 2019, 03:08
Bunuel wrote:
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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Re: In a sequence a1, a2, …, each term after the first is found by taking  [#permalink]

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12 Jun 2019, 05:35
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Top Contributor
Bunuel wrote:
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

Cheers,
Brent
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Re: In a sequence a1, a2, …, each term after the first is found by taking  [#permalink]

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17 Jun 2019, 17:48
1
Bunuel wrote:
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

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Re: In a sequence a1, a2, …, each term after the first is found by taking   [#permalink] 17 Jun 2019, 17:48
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