The 6th term from the end of the G.P. 8,4,2,1......,1/1,024 is

Correct Answer is : 132\;1/32321

Correct Answer is : 132\;1/32321

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Sequence and Series

Question : 35 of 86

Marks: +1, -0

8,4,2,1......,\frac{1}{1,024}

Solution:

Concept:

Find the total number of terms, then locate the 6th term from the end by reversing the order.

Explanation:

Given GP:

8, 4, 2, 1, \ldots, \frac{1}{1024}

with common ratio

r = \frac{1}{2}

Last term

a_n = \frac{1}{1024}

. Using

a_n = a \cdot r^{n-1}

\frac{1}{1024} = 8 \cdot \left(\frac{1}{2}\right)^{n-1}

Simplify:

\frac{1}{1024} = \frac{8}{2^{n-1}} \Rightarrow 2^{n-1} = 8 \times 1024 = 8192 = 2^{13}

Thus

n-1 = 13

, so

n = 14

terms.

The

k

-th term from the end is the

(n-k+1)

-th term from the start. For

k=6

, it is the

(14-6+1)=9

-th term.

9

-th term:

a \cdot r^{8} = 8 \cdot \left(\frac{1}{2}\right)^8 = \frac{8}{256} = \frac{1}{32}

Answer:

\frac{1}{32}

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