Question

The G.P. whose 3rd and 6th terms are 1, –1/8 respectively is

Answer 1

Answer:

$third \: term = 1 \\ a {r}^{2} = 1 \\ sixth \: term = \frac{ - 1}{8} \\ a {r}^{5} = \frac{ - 1}{8} \\ dividing \: both \: of \: them \: \\ \frac{a {r}^{5} }{a {r}^{2} } = \frac{ \frac{ - 1}{8} }{1} \\ {r}^{3} = \frac{ - 1}{8} \\ {r}^{3} = \frac{ - 1}{ {2}^{3} } \\ {r}^{3} = - {2}^{ - 3} \\ r = \frac{1}{2}$

$a {r}^{2} = 1 \\ a { \frac{1}{2} }^{2} = 1 \\ \frac{a}{4} = 1 \\ a = 4$

now from a and d we can find the gp