Question

Find the 4th term from the end of the series 8,4,2,...........,1/128

Answer 1

First term, t1 = 8
Common ratio, r = 1/2
Let number of terms be n
Last term, tn = 1/128
$tn = t1 \times {r}^{n - 1} \\ \frac{1}{128} = 8 \times {( \frac{1}{2}) }^{n - 1} \\ {( \frac{1}{2}) }^{n - 1} = \frac{1}{128 \times 8} = {( \frac{1}{2}) }^{10} \\ n - 1 = 10 \\ n = 11$

So, 4th term from end is 8th term from beginning.
$t8 = 8 \times {( \frac{1}{2} )}^{8 - 1} = 8 \times \frac{1}{ {2}^{7} } = \frac{1}{16}$

Answer 2

Answer:

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