In the sequence of non-zero numbers t1, t2,t3,......tn,.....

correction: tn+1 =tn/2

1) SUFFICIENT: t3=1/4 hence t4=1/4/2=1/8 and t5=1/8/2=1/16
2) SUFFICIENT: t1-(t1/2/2/2/2)=15/16 -> t1-(t1/16)=15/16 -> t1=1


If anybody has questions can write to me: laura81@blu.it (I teach GMAT in MILANO)

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.
OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-sequence-1-2-4-8-16-32-each-term-after-the-104175.html