Question

The sum of the first three terms of a G.P. is $\dfrac{{13}}{{12}}$ and their product is $ - 1$. Find G.P.

Answer 1

Hint: Let the terms of the G.P. be- $\dfrac{a}{r},a,ar$ where the common ratio is r. Add the terms and equate to$\dfrac{{13}}{{12}}$and multiply the terms and equate to $ - 1$. Find the value of a and put in the sum of the first three terms. A quadratic equation will be formed, solve it to find the value of r. Then put the value of a and r in $\dfrac{a}{r},a,ar$ to get the answer.

Complete step-by-step answer:
Given, the sum of the first three terms of G.P. = $\dfrac{{13}}{{12}}$
The product of the first three terms = $ - 1$
We have to find the G.P.
We know that in G.P. there is a common ratio between the consecutive terms of the series.
Let the three terms of G.P. be $\dfrac{a}{r},a,ar$ where the common ratio is r.
Then according to the question,
$ \Rightarrow \dfrac{a}{r} + a + ar = \dfrac{{13}}{{12}}$ - (i)
And $a \times ar \times \dfrac{a}{r} = - 1$ - (ii)
Then on solving eq. (ii) we get,
$ \Rightarrow {a^3} = - 1$
So we get,
$ \Rightarrow a = - 1$
Now on substituting the value of ‘a’ in eq. (ii) we get,
$ \Rightarrow \dfrac{{ - 1}}{r} - 1 - r = \dfrac{{13}}{{12}}$
On taking LCM, we get-
$ \Rightarrow \dfrac{{ - 1 - r - {r^2}}}{r} = \dfrac{{13}}{{12}}$
On cross-multiplication, we get-
$ \Rightarrow 12\left( { - 1 - r - {r^2}} \right) = 13r$
On solving we get,
$ \Rightarrow - 12 - 12r - 12{r^2} = 13r$
On multiplying negative sign both side we get,
$ \Rightarrow 12 + 12r + 12{r^2} = - 13r$
On simplifying we get,
$ \Rightarrow 12 + 12r + 13r + 12{r^2} = 0$
Now addition and rearranging the terms we get,
$ \Rightarrow 12{r^2} + 25r + 12 = 0$ - (iii)
This equation is in a quadratic form so we can factorize it to find the value of r.
On factorizing we get,
$ \Rightarrow 12{r^2} + 16r + 9r + 12 = 0$
On simplifying we get,
$ \Rightarrow 4r\left( {3r + 4} \right) + 3\left( {3r + 4} \right) = 0$
On taking $\left( {3r + 4} \right)$ common, we get-
$ \Rightarrow \left( {4r + 3} \right)\left( {3r + 4} \right) = 0$
$ \Rightarrow \left( {4r + 3} \right) = 0{\text{ or }}\left( {3r + 4} \right) = 0$
$ \Rightarrow r = \dfrac{{ - 3}}{4}$ or $r = - \dfrac{4}{3}$
On substituting the value of ‘a’ and ‘r’ in the value of three terms, we get-
$ \Rightarrow \dfrac{3}{4}, - 1,\dfrac{4}{3}$ or $\dfrac{4}{3}, - 1,\dfrac{3}{4}$
Hence the G.P. is- $\dfrac{3}{4}, - 1,\dfrac{4}{3}$ or $\dfrac{4}{3}, - 1,\dfrac{3}{4}$.

Note: Here we can also use the method of discriminant to find the value of r. The formula used to find the value of x for quadratic equation $a{x^2} + bx + c = 0$ is given as-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
On comparing the equation (iii) with standard quadratic equation we get,
$ \Rightarrow $ a=$12$ , b=$25$ , c=$12$ and x=r
On applying the formula, we get-
$ \Rightarrow $ r=$\dfrac{{ - 25 \pm \sqrt {{{25}^2} - 4 \times 12 \times 12} }}{{2 \times 12}}$
On solving we get,
$ \Rightarrow $ r=$\dfrac{{ - 25 \pm \sqrt {625 - 576} }}{{24}} = \dfrac{{ - 25 \pm 7}}{{24}}$
$ \Rightarrow $ r=$\dfrac{{ - 25 + 7}}{{24}}$ or r=$\dfrac{{ - 25 - 7}}{{24}}$
On solving we get,
$r = \dfrac{{ - 18}}{{24}} = \dfrac{{ - 3}}{4}$ or r=$\dfrac{{ - 32}}{{24}} = \dfrac{{ - 4}}{3}$