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

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

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 \(a_1\)

My answer:

Sum = (Avrg)(N)

27/3 = Avrg = 9

Since 9 = N3

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

D

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

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

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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