Question

The 13th term of a G.P are respectively equal to 24 and 3/16 find the series and 25th term

Answer 1

$a {r}^{n - 1} = an$

6th term is 24

$a {r}^{6 - 1} = 24 \\ \\ a {r}^{5} = 24........(1) \\ \\ a {r}^{13 - 1} = \frac{3}{16} \\ \: \\ a {r}^{12} = \frac{3}{16} ........(2) \\ \\ divided \: (2) \: by \: (1) \\ \\ \frac{a {r}^{12} }{a {r}^{5} } = \frac{ \frac{3 }{16} }{24} \\ \\ {r}^{7 } = \frac{3}{16} \times \frac{1}{24} \\ \\ {r}^{7} = \frac{1}{128} \\ \\ {r}^{7} = ( { \frac{1}{2} })^{7} \\ \\ r = \frac{1}{2}$

put this value of r in (1)

$a ({ \frac{1}{2} })^{5} = 24 \\ \\ a = 24 \times {2}^{5} \\ \\ a = 768$

hence the series is 768,384,192,128,64,24.........