Question

The sum of first three terms of a G.P is 13/12 and their product is -1 Find the common ratio and the terms​

Answer 1

Step-by-step explanation:

$let \: the \: three \: terms \: of \: g \: p \: are \: \: \: \: \ \\ \frac{a}{r} \: .\: \: a \: \: . \: ar \\ according \: to \: question \\ \frac{a}{r} + a + ar = \frac{13}{12} ..........(1)\\ \frac{a}{r} \times a \times ar = - 1 \\ {a}^{3} = - 1 \\ a = - 1$

$put \: the \: value \: of \: a \: in \: equation \: (1). \\ \frac{( - 1)}{r} + ( - 1) + ( - 1) \times r = \frac{13}{12} \\ \frac{ - 1}{r} - 1 - r = \frac{13}{12} \\ \frac{ - 1 - r - {r}^{2} }{r} = \frac{13}{12} \\ - 12 - 12r - 12 {r}^{2} = 13r \\ 12 {r}^{2} + 12r + 13r + 12 = 0 \\ 12 {r}^{2} + 25r + 12 = 0$

$12 {r}^{2} + 16r + 9r + 12 = 0 \\ 4r(3r + 4) + 3(3r + 4) = 0 \\ (4r + 3)(3r + 4) = 0 \\ r = \frac{ - 3}{4} \: \: \: \: \: . \: \: \: \frac{ - 4}{3}$

$terms = \frac{4}{3} . - 1. \frac{3}{4} \: \: and \\ \frac{3}{4} . - 1. \frac{4}{3}$