Question

find 9th term of GP√2,1/√2,1/2√2.......
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Answer 1

Answer:

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Step-by-step explanation:

Tn of this GP =

${ar}^{n}$

where a=√2 and r=1/2

$tn = \sqrt{2} ( { \frac{1}{2} }^{n} ) \\ = \frac{ \sqrt{2} }{ {2}^{n} } \\ tn= {2}^{ \frac{1}{2} - n }$

put n=9

$t9 = {2}^{ \frac{1}{2} - 9 } = {2}^{ \frac{ - 17}{2} }$