Bunuel wrote:
In an increasing sequence of 10 numbers each number is two times the previous one. If the sum of the numbers in the sequence is 3069, what is the value of the 3rd number?
A. 2
B. 3
C. 4
D. 12
E. 24
I see two ways to approach this: either by working backwards, or by representing the sequence using variables and solving with algebra.
The advantage of working backwards, is that I don't have to think about the algebra! The disadvantage is that I might end up adding ten numbers together, multiple times, which could take a while. However, I'm hoping that some kind of shortcut will show up when I start working backwards that might help me figure it out without actually doing ALL of the math. I'm going to at least try it.
Start with answer choice B. If the 3rd number in the sequence is 3, then the 2nd number is half this, or 1.5. The first number is half the second number, or 0.75.
Will the sequence sum to 3069? No, that doesn't make sense - almost all of the numbers in the sequence will be integers, but the first two will be decimals that don't add to an integer. So, the overall sum won't be an integer. Eliminate B. You can also eliminate A for the same reason! If A was the right answer, the first term in the sequence would be 0.5, and the sum wouldn't be an integer.
Next, try answer choice D. The beginning of the sequence would look like this:
3, 6, 12, 24, 48, ...
Let's clean it up by dividing each of these terms by 3:
3(1), 3(2), 3(4), 3(8), 3(16), ...
In other words, the sequence is:
3(2^0), 3(2^1), 3(2^2), 3(2^3), 3(2^4), ... 3(2^9)
Does this equal 3069? Well, 3069/3 = 1023. We want to know whether 2^0 + 2^1 + ... + 2^9 adds up to 1023.
If you happen to know this interesting math fact, you're already done: the sum of the powers of 2 is equal to the next largest power of 2, minus 1. So, the sum of the powers of 2, from 2^0 through 2^9, is equal to 2^10-1. Therefore, the sum IS 1023, so (D) is the right answer.
Or, you could estimate! (C) is only 1/3 as big as (D), and (E) is twice as big as (D), so even if you can just get moderately close to (D) by estimation, you've got it.
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