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

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Q49  V42
Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
1
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Hello, quantum,
this is my attempt to explain why it's D:

Quote:
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

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.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
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.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
Thanks for the explanation Bunuel. I can see it now clearly.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
2
Kudos
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.
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
I am having trouble understanding how you went from T3 = T2/2 = T1/4

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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
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g3lo18 wrote:
I am having trouble understanding how you went from T3 = T2/2 = T1/4

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
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
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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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
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Re: In the sequence of nonzero numbers t1, t2, t3, , tn, , tn+1 [#permalink]
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