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

Question

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

Archit3110 · Answer

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....

baru · Answer

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

EncounterGMAT · Answer

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

T1101 · Answer

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

jfranciscocuencag · Answer

My answer:

Sum = (Avrg)(N)

27/3 = Avrg = 9

Since 9 = N3

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

D

thyagi51 · Answer

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

Posted from my mobile device

BrentGMATPrepNow · Answer

By definition:
a1 = a1
a2 = a1 + c
a3 = a1 + c + c = a1 + 2c
a4 = a1 + c + c + c = a1 + 3c
a5 = a1 + c + c + c + c = a1 + 4c

GIVEN: a1+a3+a5=27  
We can write: a1 + (a1 + 2c) + (a1 + 4c) = 27
Simplify: 3a1 + 6c = 27
Divide both sides by 3 to get:  a1 + 2c = 9

What is the value of  a2+a4    
a2 + a4 = (a1 + c) + (a1 + 3c)
= 2a1 + 4c
= 2( a1 + 2c )
= 2( 9 )
= 18

Answer: D

mpobisetty · Answer

a + (a + 2d) + (a + 4d) = 27
So, 3a + 6d = 27
So, a + 2d = 9

We have to find out (a + d) + (a + 3d) = 2a + 4d = 2*(a + 2d) = 2*9 = 18

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