In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1

Question

In the sequence of nonzero numbers  t_1 ,  t_2 ,  t_3 , …,  t_n , …,  t_{n+1}=\frac{t_n}{2}  for all positive integers n. What is the value of  t_5 ?

(1)  t_3 = \frac{1}{4}

(2)  t_1 - t_5 = \frac{15}{16}

brokerbevo · Accepted Answer

see attached

Bunuel · Answer

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?

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.

greenoak · Answer

Hello, quantum, 
this is my attempt to explain why it's D:

Here we have geometric progression, i.e. series where t2=t1*q, t3=t2*q, …, tn+1=tn*q. In our case, q=0.5. Also note that tn+1=t1*q^n

So, basically, to answer this question, it is sufficient to know the value of any of the tn.

1) Explicitly gives us the value for t3, so it’s sufficient.

2) So, let’s see if we can obtain the value of t1 from this statement, using the formula tn+1=t1*q^n:

15/16=t1-t5=t1-t1*q^4 = t1*(1-q^4) = t1*(1-1/16) = t1*15/16.

So, from here it follows that t1=1 and t5=t1*q^4 = 1*1/16 = 1/16.

Sufficient.

I hope that helped.

gettinit · Answer

Good explanations but I got confused at how you equated tn+1=tn/2? I thought the it was the entire expression that equaled to tn/2? sorry but I am a bit confused. Thanks.

gettinit · Answer

Thanks for the explanation Bunuel. I can see it now clearly.

ravishankar1788 · Answer

statement(1)

tn+1 = tn/2 
t3 = 1/4 
t4 =t3/2 = 1/8 
t5 = t4/2 = 1/16. Sufficient.

statement (2) 
t1 - t5 = 15/16 
t1 = t5 + 15/16 = (16t5 + 15) /16 
t2 = t1/2 = (16t5 + 15) / 32 
t3 = t2/2 = (16t5 + 15) /64 
t4 = t3/2 = (16t5 + 15) /128 
t5 = t4/2 = (16t5 + 15) /256

256t5 = 16t5 + 15 
240t5 = 15 
t5 = 15/240 = 5/80 = 1/16 
Sufficient.

Thus, D.

g3lo18 · Answer

I am having trouble understanding how you went from T3 = T2/2 = T1/4

Please explain

abhimahna · Answer

Notice that we are given T(n+1) = T(n)/2

So, substitute n = 2, you will get T(3) = T(2)/2.

Now substitute n =1 , you will get T(2) = T(1)/2

So, we can say T(3) = T(2)/2 = T(1)/4

g3lo18 · Answer

OMG I see it now. Thank you. Wow these are so annoying to deal with. Guess I need to get accustomed to it.

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In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1

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