Question

the 10th term of the GP is


$1 \div 2 {}^{8}$
and the 5th term is
$1 \div 2 {}^{3}$

what is the common ratio...
​

Answer 1

Answer:

r = 1/2

Step-by-step explanation:

Let First Term be 'a' and Common Ratio be 'r'

$ATQ, \\\\a_{10} = \frac{1}{2^{8} } = \frac{1}{256} = ar^{9} \\\\a_{5} = \frac{1}{2^{3} } = \frac{1}{8} = ar^{4}$

$.: a = \frac{1}{256 X r^{9} } = \frac{1}{8 X r^{4} } \\\\Hence, \\\\\frac{1}{256 X r^{9} } = \frac{1}{8 X r^{4} } \\\\\frac{r^{4} }{r^{9} } = \frac{256}{8} \\r^{4-9} = 32\\r^{-5} = 2^{5} \\r = 2x^{\frac{5}{-5} } \\r = 2^{-1} \\\\.: Ratio = r = \frac{1}{2}$

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