Math, asked by aldrin9942, 6 hours ago

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

Answers

Answered by shrirampawar249
3

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

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