Source: Official Guide Revised GRE 1st Ed. Part 8; Section 5; #24

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# In the sequence above, each term after

In the sequence above, each term after the first term is equal to the preceding term plus the constant c. If a1 + a3 + a5 = 27, what is the value of a2 + a4? a2 + a4 =

### 3 Explanations

3

Jacob Elder

In the explanation, I see as one example, you determine that a(1) is 8, a(2) is 9, and a(3) is 10. However the question defines the terms as "equal to the preceding term plus the constant c". It seems that you just solved for the terms, but don't we need to solve for the constant too? Is it not important to the question to know what each term has added to it, but instead, look at what the sum is after the constant is added?

Jan 24, 2016 • Comment

Cydney Seigerman, Magoosh Tutor

Hi Jacob :)

As we can see through the various examples in the explanation video, the constant c itself can vary. It's important to keep in mind that c doesn't have to be an integer. In the example you've mentioned, we have the terms

a_1 = 8
a_3 = 9
a_5 = 10

These terms satisfy the equation a_1 + a_3 + a_5 = 27. We can solve for c in this case by recognizing that the difference between every 2 terms is 1 (e.g. a_3 - a_1 = 9-8 = 1). So, c, the difference between every term, is half of that or 1/2:

a_1 = 8
a_2 = 8.5
a_3 = 9
a_4 = 9.5
...

The sum a_2 + a_4 = 8.5 + 9.5 = 18.

This is one possible sequence that satisfies the conditions of the problem. Another possible set of terms is

a_1 = 3
a_3 = 9
a_5 = 15

The sum of these three terms is still 27 but in this case, c = 3. So, a_2 = 6 and a_4 = 12 and the sum a_2 + a_4 = 18, which is what we got when c = 0.5.

The constant itself is not important but rather the fact that the terms satisfy the restriction of the question.

I hope this helps :)

11

The way I found the solution to this question was by finding the value of "a" first: (using the sequence rule) (a+1)+(a+3)+(a+5)=27----> 3a+9=27, so a= 6. Then I use the 6 to substitute the a's in the equation, a2+a4---> (6+2)+(6+4)=18, Is this approach correct?

Oct 13, 2013 • Comment

Eric Kuzmenko

a1+a3+a5 = a1+(a1+2c)+(a1+4c) = 3a1+6c = 27
a2+a4 = (a1+c)+(a1+3c) = 2a1+4c = 2/3(3a1+6c) = 2/3(27) = 18

Jose Veloz

Hi Eric, I didn't get how you came up with "2/3(3a1+6c)"

Thanks!

Cydney Seigerman, Magoosh Tutor

Hi Jose!

What Eric has done is take the expression (2a1 + 4c) and "factor out" 2/3 from both terms. To do this, he divided each term by 2 and multiplied each term by 3:

2a1 = 2/3*(3a1)
4c = 2/3*(6c)

If we were to multiply out the right side of the equations above, we would see that we are left with the original term :) The reason Eric factored out (2/3) from the two terms is to end up with the expression (3a1+6c), since we know the value of this expression is 27.

Hope this clears up that step!

Jose Veloz

Hi Cydney! Yes, thanks for your kind response. A nice, useful strategy to come up with the same expression.
Bests

Cydney Seigerman, Magoosh Tutor

You're very welcome, Jose! I'm glad to have helped :) Happy studying :D

6 Chris Lele, Magoosh Tutor

Oct 11, 2012 • Comment