Question

which term of g. p 1,√3,3 will be 81.

Answer 1

$a = 1 \\ r = \sqrt{3} \\ a_{n} = a {r}^{n - 1} \\ 81 = { (\sqrt{3}) }^{n - 1} \\ {3}^{4} = {3}^{ \frac{n - 1}{2} } \\ 4 = \frac{n - 1}{2} \\ n = 9$
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Answer 2

Answer:

81=1(√3)n-1

81=(√3)n-1

81/√3=n-1

81/9=n-1

9=n-1

9+1=n

10=n