Question

If the 2nd term of a gp is 2/3 and its 5th term is 16/81. find the 7th term

Answer 1

$t2 = \frac{2}{3} \\ ar = \frac{2}{3} - - - (1) \\ t5 = \frac{16}{81} \\ a {r}^{4} = \frac{16}{81} - - - (2)$

$\frac{(2)}{(1)} \: \: gives \\ {r}^{3} = \frac{8}{27} \\ r = \frac{2}{3} \\ substituting \: \: in \: \: (1) \\ a = 1$

$t7 = a {r}^{6} = 1 \times {( \frac{2}{3}) }^{6} = \frac{64}{729 }$