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Question: The sum of the first three terms of a G.P. is \(\dfrac{{13}}{{12}}\) and their product is \( - 1\). ...

The sum of the first three terms of a G.P. is 1312\dfrac{{13}}{{12}} and their product is −1 - 1. Find G.P.

Explanation

Solution

Let the terms of the G.P. be- ar,a,ar\dfrac{a}{r},a,ar where the common ratio is r. Add the terms and equate to1312\dfrac{{13}}{{12}}and multiply the terms and equate to −1 - 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 ar,a,ar\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. = 1312\dfrac{{13}}{{12}}
The product of the first three terms = −1 - 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 ar,a,ar\dfrac{a}{r},a,ar where the common ratio is r.
Then according to the question,
⇒ar+a+ar=1312\Rightarrow \dfrac{a}{r} + a + ar = \dfrac{{13}}{{12}} - (i)
And a×ar×ar=−1a \times ar \times \dfrac{a}{r} = - 1 - (ii)
Then on solving eq. (ii) we get,
⇒a3=−1\Rightarrow {a^3} = - 1
So we get,
⇒a=−1\Rightarrow a = - 1
Now on substituting the value of ‘a’ in eq. (ii) we get,
⇒−1r−1−r=1312\Rightarrow \dfrac{{ - 1}}{r} - 1 - r = \dfrac{{13}}{{12}}
On taking LCM, we get-
⇒−1−r−r2r=1312\Rightarrow \dfrac{{ - 1 - r - {r^2}}}{r} = \dfrac{{13}}{{12}}
On cross-multiplication, we get-
⇒12(−1−r−r2)=13r\Rightarrow 12\left( { - 1 - r - {r^2}} \right) = 13r
On solving we get,
⇒−12−12r−12r2=13r\Rightarrow - 12 - 12r - 12{r^2} = 13r
On multiplying negative sign both side we get,
⇒12+12r+12r2=−13r\Rightarrow 12 + 12r + 12{r^2} = - 13r
On simplifying we get,
⇒12+12r+13r+12r2=0\Rightarrow 12 + 12r + 13r + 12{r^2} = 0
Now addition and rearranging the terms we get,
⇒12r2+25r+12=0\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,
⇒12r2+16r+9r+12=0\Rightarrow 12{r^2} + 16r + 9r + 12 = 0
On simplifying we get,
⇒4r(3r+4)+3(3r+4)=0\Rightarrow 4r\left( {3r + 4} \right) + 3\left( {3r + 4} \right) = 0
On taking (3r+4)\left( {3r + 4} \right) common, we get-
⇒(4r+3)(3r+4)=0\Rightarrow \left( {4r + 3} \right)\left( {3r + 4} \right) = 0
⇒(4r+3)=0 or (3r+4)=0\Rightarrow \left( {4r + 3} \right) = 0{\text{ or }}\left( {3r + 4} \right) = 0
⇒r=−34\Rightarrow r = \dfrac{{ - 3}}{4} or r=−43r = - \dfrac{4}{3}
On substituting the value of ‘a’ and ‘r’ in the value of three terms, we get-
⇒34,−1,43\Rightarrow \dfrac{3}{4}, - 1,\dfrac{4}{3} or 43,−1,34\dfrac{4}{3}, - 1,\dfrac{3}{4}
Hence the G.P. is- 34,−1,43\dfrac{3}{4}, - 1,\dfrac{4}{3} or 43,−1,34\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 ax2+bx+c=0a{x^2} + bx + c = 0 is given as-
⇒x=−b±b2−4ac2a\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
On comparing the equation (iii) with standard quadratic equation we get,
⇒\Rightarrow a=1212 , b=2525 , c=1212 and x=r
On applying the formula, we get-
⇒\Rightarrow r=−25±252−4×12×122×12\dfrac{{ - 25 \pm \sqrt {{{25}^2} - 4 \times 12 \times 12} }}{{2 \times 12}}
On solving we get,
⇒\Rightarrow r=−25±625−57624=−25±724\dfrac{{ - 25 \pm \sqrt {625 - 576} }}{{24}} = \dfrac{{ - 25 \pm 7}}{{24}}
⇒\Rightarrow r=−25+724\dfrac{{ - 25 + 7}}{{24}} or r=−25−724\dfrac{{ - 25 - 7}}{{24}}
On solving we get,
r=−1824=−34r = \dfrac{{ - 18}}{{24}} = \dfrac{{ - 3}}{4} or r=−3224=−43\dfrac{{ - 32}}{{24}} = \dfrac{{ - 4}}{3}