Question

quantos termos tem a PG (243,81,27,...,1/3 elevado a 10)?

Answer 1

$(243;\,81;\,27;...\,; \frac{1}{3^{10}})$

$a_1 = 243 = 3^5 \\ \\q = \frac{1}{3} = 3^{-1} \\ \\ a_n = \frac{1}{3^{10}} = 3^{-10} \\ \\ \\a_n = a_1\cdot q^{n-1} \\ \\ 3^{-10} = 3^5\cdot (3^{-1})^{n-1} \\ \\ 3^{-10} = 3^5\cdot 3^{1-n} \\ \\ 3^{-10} = 3^{5+1-n} \\ \\ 3^{-10} = 3^{6-n} \\ \\ -10 = 6 - n \\ \\ n = 6 + 10 \\ \\ \boxed{n = 16}$