Question

if the 3rd and 7th term of a GP at 12 and 192 respectively find the 9th and 12th term of the sequence​

Answer 1

Answer:

$9th \: \: \: term = a {r}^{8} = 768 \\ \\ \\ 12th \: \: term = a {r}^{11} = 6144$

Step-by-step explanation:

$\\ a {r}^{2} = 12 \: \: \: \: \: \: \: \: \: \: \: \: a {r}^{6} = 192 \\ \\ \\ \\ solving \: these \\ \\ {r}^{4} = \frac{192}{12} \\ \\ {r}^{4} = 16 \\ \\ r = 2 \\ \\ a = \frac{12}{4} = 3 \\ \\ 9th \: \: \: term = a {r}^{8} = 3 \times {2}^{8} = 256 \times 3 = 768 \\ \\ \\ 12th \: \: term = a {r}^{11} = 3 \times {2}^{11} = 6144$