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In the sequence above, each term after the first term is equal to the

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In the sequence above, each term after the first term is equal to the  [#permalink]

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New post 03 Dec 2018, 02:57
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

85% (02:11) correct 15% (01:42) wrong based on 70 sessions

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Re: In the sequence above, each term after the first term is equal to the  [#permalink]

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New post 03 Dec 2018, 03:40
Bunuel wrote:
a1, a2, a3, ..., an

In the sequence above, each term after the first term is equal to the preceding term plus the constant c. If \(a_1+a_3+a_5=27\), what is the value of \(a_2 + a_4\) ?

A. 5
B. 9
C. 13
D. 18
E. 22


took me >2 mins to understand question ,if any one has any other solution please do provide solution:

I assumed a1=3
and c to be 3 as well
so
a1=3;a2=6;a3=9;a4=12;a5=15

a1+a3+a5=27 9 given

a1+a2+c+a4+c=27
a2+a4+2c+a1=27

since a1=c=3
so a2+a4= 27-(2*3+3)
a2+a4=27-9
a2+a4=18

IMO D....
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In the sequence above, each term after the first term is equal to the  [#permalink]

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New post 04 Dec 2018, 07:14
given a1+a3+a5 = 27
I assumed constant C = 1 , then sequence ==> 1,2,3,4,5,6,7
==> 1+3+5 = 9 ==>if I make C =3 that will satisfy given condition
then sequence becomes ==> 3,6,9,12,15,18...

finally a2+a4 = 6+12 = 18==> Option D
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Re: In the sequence above, each term after the first term is equal to the  [#permalink]

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New post 04 Dec 2018, 09:01
let a2=a1+c, a3=a1+2
a1+a1+2c+a1+4c=27
3a1+6c=27
a1+c=9...............(eq. 1)

We need to find a2+a4
a2+a4=a1+c+a1+3c= 2a1+4c= 2(a1+2c)=2(9) .....from eq. 1=18

Therefore, answer is option D
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Re: In the sequence above, each term after the first term is equal to the  [#permalink]

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New post 05 Dec 2018, 02:33
Bunuel wrote:
a1, a2, a3, ..., an

In the sequence above, each term after the first term is equal to the preceding term plus the constant c. If \(a_1+a_3+a_5=27\), what is the value of \(a_2 + a_4\) ?

A. 5
B. 9
C. 13
D. 18
E. 22


Took me quite some time to figure out how to get to an answer, since I was not able to calculate a definite value for \(a_1\)

\(a_1+a_3+a_5=27\) \(-->\) \(a_3=a_1+2c\) and \(a_5=a_1+4c\)

Rewrite the equation: \(a_1+a_1+2c+a_1+4c=27\) \(-->\) \(a_1+2c=9\) At this points I was quite stuck... But I then assumed integer values for \(c\)

\(c=0\) \(-->\) \(a_1=9\) \(-->\) \(a_2 + a_4=9+9=18\)
\(c=1\) \(-->\) \(a_1=7\) \(-->\) \(a_2 + a_4=8+10=18\)
\(c=2\) \(-->\) \(a_1=5\) \(-->\) \(a_2 + a_4=7+11=18\)
\(c=3\) \(-->\) \(a_1=3\) \(-->\) \(a_2 + a_4=6+12=18\)
\(c=4\) \(-->\) \(a_1=1\) \(-->\) \(a_2 + a_4=5+13=18\)

As we can see any value of \(a_1\) will result in \(a_2 + a_4=18\), hence D.

Bunuel is there any faster way to solve this question? I was quite stuck thinking i was doing something wrong, because I could not solve for a value of \(a_1\)
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In the sequence above, each term after the first term is equal to the  [#permalink]

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New post 20 Jan 2019, 18:01
My answer:

Sum = (Avrg)(N)

27/3 = Avrg = 9

Since 9 = N3

N2 + N4 MUST be 18 since (N2 + N4)/2 = N3 = 9

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Re: In the sequence above, each term after the first term is equal to the  [#permalink]

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New post 22 Jan 2019, 05:27
I started putting in answer options .. for example my reduced equation was 2a1+4c...
If 2a1+ 4c = 18
Then A1+2c= 9...
Therefore answer option d is correct

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Re: In the sequence above, each term after the first term is equal to the   [#permalink] 22 Jan 2019, 05:27
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