In an increasing sequence of 10 numbers each number is two times the p

Question

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

Chethan92 · Answer

Let the sequence be - a 2a 4a 8a 16a 32a....512a (10 numbers)

Given, Sum of these = 3069
Sum = 1023a = 3069
a = 3

Third number = 4a = 12

D is the answer.

ccooley · Answer

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.

CareerGeek · Answer

Each number is two times the previous one.
—> The series is G.P

Sum of n terms of GP = a(r^n - 1)/(r - 1) = 3069

Given, r = 2 and n = 10
—> a(2^10 - 1)/(2 - 1) = 3069
—> a(1023) = 3069
—> a = 3.

IMO Option B.

Pls Hit kudos if you like the solution

Posted from my mobile device

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

In an increasing sequence of 10 numbers each number is two times the p

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In an increasing sequence of 10 numbers each number is two times the p

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards