In the sequence of non-zero numbers a1, a2, a3, ... an, ..., the value

Question

In the sequence of non-zero numbers  a_1 ,  a_2 ,  a_3 , ...  a_n , ..., the value of  a_{(n+1)} = \frac{a_n}{2} , for all positive integers n. What is the value of  a_6 ?

(1)  a_2 =\frac{1}{2}

(2)  a_2 − a_6 = \frac{15}{32}

GMATinsight · Answer

Question : What is the value of  a_6 ?

Given :  a_{(n+1)} = \frac{a_n}{2} 
i.e. every Next term of the sequence is Half the previous term 
ie. in order to answer this question we need any one term of the series and we can find all other terms

Question REPHRASED : What is the Numerical Value of any term of the series?

Statement 1 :  a_2 =\frac{1}{2}

SUFFICIENT

Statement 2 :  a_2 − a_6 = \frac{15}{32}

Using the series  a_2  may be translated into  a_6  and then we can find the numerical value using this statement hence

a_3 = a_2 / 2 
 a_4 = a_3 / 2 = a_2/4 
 a_5 = a_4 / 2 = a_2/8 
 a_6 = a_5 / 2 = a_2/16 
i.e.  a_2 = 16*a_6

Now,  a_2 − a_6 = \frac{15}{32} 
i.e.  16*a_6 − a_6 = \frac{15}{32} 
i.e.  15*a_6 = \frac{15}{32} 
i.e.  a_6 = \frac{1}{32}

SUFFICIENT

Answer: Option D

Salsanousi · Answer

1)  a_2 =\frac{1}{2}

a_3 = \frac{1}{2/2}  =  \frac{1}{4}

a_6 will then be in the form of \frac{1}{2^n+1}

thus  a_6 = \frac{1}{32}

sufficient.

for 2)
 a_2 = \frac{a_1}{2}

a_3 = \frac{a_2}{2}

etc...

a_6 = \frac{a_1}{2^n+1}

\frac{a_1}{2} - \frac{a_1}{32} = \frac{16*a_1}{32} - \frac{a_1}{32} = \frac{15*a_1}{32}

\frac{15*a_1}{32} = \frac{15}{32}

a_1 = 1

a_6 = \frac{a_1}{2^n+1} = \frac{1}{32}

Sufficient

Answer choice D

bumpbot · Answer

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In the sequence of non-zero numbers a1, a2, a3, ... an, ..., the value

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In the sequence of non-zero numbers a1, a2, a3, ... an, ..., the value

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