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

Given:

The sum of first three terms of a G.P = 13/12

their product =-1

To find :

The common ratio and the terms=?

Solution :

$\large \red{\bold { \underline{step \: by \: step \: explaination}}}$

$\\ \implies\green{\sf{ Let \: the \: terms \: of \: G.P. \: be \: \dfrac{a}{r} \: , \: a , \: ar }}$

$\blue\bigstar\pink{\sf{It \: is \: given \: that}}$

$\\ \implies \green{ \sf { \frac{a}{r} + a + ar = \frac{13}{12} }} \: \: and \: \: \blue{ \sf{ \: \frac{a}{r} \times a \times ar = - 1}}$

$\\ \implies \large \purple{ \sf{ \: a \bigg( \frac{ {r}^{2} + r+ 1 }{r} \bigg) = \frac{13}{12} }} \: \: and \: \purple{ \sf{ \: {a}^{3} = - 1}}$

$\\ \implies \sf \purple {\: a \: = - 1 \: \: and \: \: a( {r}^{2} + r + 1) = \frac{13r}{12} }$

$\\ \implies \sf{ \purple{ \: {r}^{2} + r + 1 = - \frac{13r}{12} }}$

$\\ \implies \sf { \purple{ \: 12 {r}^{2} + 25r + 12 = 0}}$

$\\ \implies \sf{ \purple{ \: 12 {r}^{2} + \pink{16r + 9r }+ 12 = 0}}$

$\\ \implies \sf{ \purple{(3r + 4)(4r + 3) = 0}}$

$\\ \implies \sf{ \purple{ \: \red{ r = \frac{ - 4}{3} }\: and \: \: \red{r = \frac{ - 3}{4} }}}$

$\implies \sf{ \purple{ \boxed{ \: \red{r = \frac{ - 3}{4} } \: \: \: and \: \: \: \red{ r = \frac{ - 4}{3} }}}}$

$\\ \therefore\sf { \blue{ \: hence \: three \: numbers \: are \: \frac{ - 3}{4} \: and \: - 1 \: and \: \frac{ - 4}{3} }}$