Question

Find the third term of gp whose common ratio is 3 and the sum of whose first 7 terms is 2186

Answer 1

$given \: \: r = 3 \\ s7 = 2186 \\ \frac{a( {r}^{7 } - 1) }{r - 1} = 2186 \\ a = \frac{2186 \times (3 - 1)}{ {3}^{7} - 1 } = 2 \\ t3 = a \times {r}^{3 - 1} \\ = 2 \times {3}^{2} = 18$