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

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

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New post 16 May 2018, 04:37
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

74% (01:54) correct 26% (02:05) wrong based on 54 sessions

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

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New post 16 May 2018, 12:02
Bunuel wrote:
In a sequence a1, a2,…, 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


In the sequence, since a1 = 2, a2 will be equal to (-2) + 1 = -1.
Similarly, a3 is -(-1) + 1 = 2 and a4 = -2 + 1 = -1.

We see a pattern emerge where the odd term is 2 and the even term is -1.
The sum of every pair of odd and even term is 1. The first 99 terms contain
49 pairs of even and odd terms and the 99th term.

Therefore, the sum of the first 99 terms is 49*1 + 2 = 51(Option 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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New post 16 May 2018, 12:09
Bunuel wrote:
In a sequence a1, a2,…, 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


\(a_1=2 => a_2=-2+1=-1, a_3=-(-1)+1=2, a_4=-2+1=-1, a_5=-(-1)+1=2\), and so on

So we see here all the Odd terms are same and equal to \(2\) and all the even terms are same and equal to \(-1\)

from \(1\) to \(99\), there are \(49\) even terms and \(50\) odd terms.

Hence the sum \(= 50*2+49*(-1)=51\)

Option 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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New post 18 May 2018, 11:25
Bunuel wrote:
In a sequence a1, a2,…, 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


a1 = 2

a2 = -1

a3 = 2

a4 = -1

a5 = 2

a6 = -1

We see that each odd/even pair sums to positive 1.

There are 49 even/odd pairs in the first 98 terms; hence, the sum is 49. Since a99 is an odd-numbered term, a99 = 2. So the sum of the first 99 terms is 49 + 2 = 51.

Answer: 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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New post 18 May 2018, 18:30
In a sequence a1, a2,…, 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?

A1 = 2
A2 = -2 + 1 = -1
A3 = -(-1)+1 = 2
A4 = -2+1 = -1

The pattern repeats. If you add A1+A2 you get 1, so similarly there are 49 pairs in first 98 numbers - 49 , the 99th term is 2

Ans: 49 + 2 = 51

Ans: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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New post 23 Mar 2019, 06:49
why can't use the sum of terms rule here . Sum= number ( first + last) / 2 , instead of checking the pattern.
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Re: In a sequence a1, a2,…, each term after the first is found by taking   [#permalink] 23 Mar 2019, 06:49
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