In the sequence 1, 2, 4, 8, 16, 32, …, each term after the

This is the concept of a geometric progression (GP). In a GP, every term is related to the previous term by a fixed ratio r.
So \(t_2 = t_1*r\)
\(t_3 = t_2*r = t_1*r*r\) (substituting from above)
\(t_4 = t_3*r = t_1*r*r*r\)
I hope you understand this.

Here, you are given that \(t_{n+1} = t_n *(\frac{1}{2})\)
So r = 1/2
\(t_2 = t_1 * (1/2)\)
\(t_3 = t_1*(1/2)*(1/2)\)
\(t_4 = t_1 *(1/2)^3\)
\(t_5 = t_1 * (1/2)^4\)

Statement 1 gives you \(t_1\) so you get \(t_5\) easily.
Statement 2 gives you \(t_1 - t_5\). You can substitute \(t_5\) from above and get \(t_1\) and then \(t_5\).

Check out this post on GPs: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/04 ... gressions/
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Given: \(t_{n+1}=\frac{t_n}{2}\). So \(t_2=\frac{t_1}{2}\), \(t_3=\frac{t_2}{2}=\frac{t_1}{4}\), \(t_4=\frac{t_3}{2}=\frac{t_1}{8}\), ...

Basically we have geometric progression with common ratio \(\frac{1}{2}\): \(t_1\), \(\frac{t_1}{2}\), \(\frac{t_1}{4}\), \(\frac{t_1}{8}\), ... --> \(t_n=\frac{t_1}{2^{n-1}}\).

Question: \(t_5=\frac{t_1}{2^4}=?\)

(1) \(t_3=\frac{1}{4}\) --> we can get \(t_1\) --> we can get \(t_5\). Sufficient.
(2) \(t_1-t_5=2^4*t_5-t_5=\frac{15}{16}\) --> we can get \(t_5\). Sufficient.

Answer: D.

Generally for arithmetic (or geometric) progression if you know:

- any particular two terms,
- any particular term and common difference (common ratio),
- the sum of the sequence and either any term or common difference (common ratio),

then you will be able to calculate any missing value of given sequence.

Hope it helps.
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Thanks Bunuel for the detailed explanation.
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In the sequence of nonzero numbers t1, t2, t3, …, tn, …, tn+1 = tn / 2 for all positive integers n. What is the value of t5?
(1) t3 = 1/4
(2) t1 - t5 = 15/16

I sloved this question, but I heard antoher theory that can solve this question.
The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1
And I don't understand how this works.
Can some explain how this concept work??

Hi, it is very simple.

The theory is that tn+1 = tn / 2 equals tn=t1*(1/2)^n-1. In this theory they just generalize it.

Given -- tn+1 = tn/2

Putting n = n- 1

tn= t(n-1) *1/ 2 ----- consider this first equation.
while

t(n-1) = t(n-2)*1/2 ----- consider this 2nd equation.

if you put t(n-1) in first equation

tn = t(n-2) * 1/ 2^2

so when we need t1 on the right hand side.

tn=t1*(1/2)^n-1

Easy way to understand is ---

t2 = t1/2

t3 = t2 / 2

t3 = t1* 1/2^2

So on generalization

tn = t1* 1/ 2^(n-1)

Hope I make it clear.

Thanks! I considered your explanation and got the idea.

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