The number in the series are the average of the numbers before and after so the difference between two consecutive numbers in the odd or even place will always be the same.
For eg : 1,5,9,13,17,21,25,29 .....The difference between successive numbers in odd or even place will always be the same. That is 8.

So, the answer is "C".

Can someone please help with this question?

Regards
Arorni

What is unclear sir ??

Thanks for replying back Carcass. I am trying to solve it algebraically, but each time somehow messing it up. What should be the right approach to solve this question quickly? Can you help me with the algebraic solution to this question?

If manipulate the two quantities

\(a^10-a^8 >>> a^8(a^2-1)\)

\(a^5-a^3 >>> a^3(a^2-1)\)

Considering that this is an arithmetic progression the key element is always the same \((a^2-1)\)

which means that the distance between the terms \(a^8\) and \(a^3\) is always the same.

So the two quantities must be equal

For a sequence problem, the quickest approach is to build the sequence with real numbers. The algebraic approach is often too complex or time-consuming to grasp.

Let, the first number, a1 = 0.
Now, let's put the next number, a2 = 1.
We choose 0 and 1 as the first two numbers as this makes calculations easy.
According to the question, a2 is the average of a1 and a3. We have our a1 and a2. Now for a3, ask "1 is the average of 0 and WHAT number?"
The answer is 2, as (2+0)/2 =1.
a3 = 2
So, our sequence is 0,1,2 so far.
For a4, we ask the question, "2 is the average of 1 and WHAT number?" well, that's 3.
a4=3
Now we already see the pattern, it's 0,1,2,3,4,5,6,7,8...

We can easily compare Quantity A and Quantity B using this sequence. It's quite obvious now the Answer will be C.

Hope this helps!

Wowww.. that's great! a nice and clean way to solve this problem.

Thanks,
arorni

Thanks for replying back Saumik. You are right. I could have tried forming the sequence using simple numbers. It's just that I have a habit to solve everything algebrically. I will keep that in mind next time.

Your approach must be flexible.

Your goal is to solve correctly the question in the shortest amount of time.

Which strategy is best you should use.

The more approaches you know the higher is the probability to nail.the question fast

Regards

Posted from my mobile device

Carcass sir, I do not understand this logic sir.

\(a^10-a^8 >>> a^8(a^2-1)\)

This piece in red is like a constant X

Two numbers which are a^8 and a^3 are multiplied by the same number

a^8 and a^3 * x=1 for instance or x=2 or x=3

the distance between a^8 and a^3 must be constant

Thank you

Sorry, but I don't understand - firstly, these are not exponents but sequence placeholders - a8, a10. And If 2 different placeholders a8 and a3 are multiplied by a2-1 i.e. second term -1 then how can we say Qa and Qb are equal algebraically?

I have one thing to perhaps humbly add to the very excellent answers and methods already offered here. Here's what I did:
1. an = (an - 1 + an + 1)/2
2. 2an = an - 1 + an + 1
3. Realize that any number in the sequence, when doubled, has to equal the value before it plus the value after it.
4. This can be discovered by plugging in numbers:
If a4 = 4, then 2(4) = 8, so term 3 and term 5 have to add up to 8.
If a4 = 8, then 2(8) = 16, so term 3 and term 5 have to add up to 16.

Let's assume that a4 = 4. That means that a3 + a5 = 2(4) = 8. Just to experiment, what if we assume that term 3 is 3 and term 5 is 5? Then, term 5 should look like this:
2(a5) = a4 + a6
2(5) = 10 = 4 + a6
6 = a6

That means that a4 = 4, a5 = 5, and a6 = 6. That means an = n.

Then, option A is simply 10 - 8 = 2 and option B is also 2.

Hence, C.

Well for every set of consecutive numbers, it is true that "each term after the first is equal to the average of the preceding term and the following term."

1,2,3,4,5,6,7,8,......n
2,4,6,8,10,12,........n
5,10,15,20,25,.......n

It's just another way of saying that the sequence is a set of consecutive numbers.

Thus the difference between two values stays the same.

Answer C

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