Question

Find the common ratio of the certain series in Geometric Progression i.e. G.P., if the sum of its three terms is $ 19 $ and the product is $ 216 $ respectively.
(a) $ \dfrac{3}{2} $
(b) $ \dfrac{2}{3} $
(c) None of these
(d) Cannot be determined

Answer 1

Hint: The given problem revolves around the concepts of solving the sequence of Geometric Progression (G.P.) whose common difference is the inverse proportional in the respective sequence in G.P. Then, by considering the certain conditions, the individual terms can be calculated along with common differences particularly. Use factorization formulae to get the desired value.

Complete step-by-step answer:
The given condition is about Geometric Progression noted as G.P. respectively.
$a,ar,a{r^2}$be the required sequence in an G.P.
Where, ‘$a$’ is any term, number, integer, variable, etc. and ‘$r$’ is its common difference in G.P. respectively
As a result, by the first condition,
$a + ar + a{r^2} = 19$ … (i)
Similarly,
By the second condition, we get
$a \times ar \times a{r^2} = 216$
Solving the equation mathematically,
${a^3}{r^3} = 216$
Taking cube roots on both the sides, we get
$ar = 6$
$a = \dfrac{6}{r}$ … (ii)
Substituting equation (ii) in the equation (i), we get
$\dfrac{6}{r} + \dfrac{6}{r}r + \dfrac{6}{r}{r^2} = 19$
\[\dfrac{6}{r} + 6 + 6r = 19\]
Dividing the equation by ‘r’, we get
\[6 + 6r + 6{r^2} = 19r\]
\[6 + 6r - 19r + 6{r^2} = 0\]
Solving the equation mathematically, we get
\[6{r^2} - 13r + 6 = 0\]
As a result, the quadratic equation is formed of degree $2$
Hence, solving it by the factorization formula, we get
\[r = \dfrac{{ - \left( { - 13} \right) \pm \sqrt {{{\left( { - 13} \right)}^2} - 4\left( 6 \right)\left( 6 \right)} }}{{2 \times 6}}\]
Solving the equation mathematically, we get
\[r = \dfrac{{13 \pm \sqrt {169 - 144} }}{{12}}\]
\[r = \dfrac{{13 \pm \sqrt {25} }}{{12}}{\text{ }} = {\text{ }}\dfrac{{13 \pm 5}}{{12}}\]
Since, equation may contain two values, as a result solving it predominantly, we get
\[r = \dfrac{{13 + 5}}{{12}}{\text{ or }}\dfrac{{13 - 5}}{{12}}\]
\[r = \dfrac{{18}}{{12}}{\text{ or r = }}\dfrac{8}{{12}}\]
Converting the equation in simplest form by dividing the equation by six and four respectively, we get
\[r = \dfrac{3}{2}{\text{ or r = }}\dfrac{2}{3}\]
Since, substituting either in equation (i) or (ii), we will get that both the value satisfies that equations i.e. say, we will substitute \[r = \dfrac{3}{2}{\text{ and r = }}\dfrac{2}{3}\] in equation (i), we get
At first, when $r = \dfrac{3}{2}$
Equation (i) becomes,
$a + a\dfrac{3}{2} + a\dfrac{9}{4} = 19$
$4a + 6a + 9a = 76 = 19a$
Hence,
$a = 4$
When $r = \dfrac{2}{3}$
Equation (ii) becomes,
$a + a\dfrac{2}{3} + a\dfrac{4}{9} = 19$
Hence,
$9a + 6a + 4a = 171$
$19a = 171$
$a = 9$
Now, hence, at $a = 4,{\text{ }}r = \dfrac{3}{2}$
Equation (i) becomes,
$4 + 4 \times \dfrac{3}{2} + 4 \times \dfrac{9}{4} = 4 + 6 + 9{\text{ }} = {\text{ 19}}$ … (iii)
Similarly,
At $a = 9,{\text{ }}r = \dfrac{2}{3}$
Equation (i) becomes,
\[9 + 9 \times \dfrac{2}{3} + 9 \times \dfrac{4}{9} = 9 + 6 + 4{\text{ }} = {\text{ 19}}\] … (iv)
As a result, from (i), (iii) and (iv), we get
It seems that, both the values for ‘r’ satisfy the given conditions.
$ \Rightarrow \therefore $The option (c) is correct!
So, the correct answer is “Option c”.

Note: One must be able to know the basic idea behind these type sequences such as Arithmetic Progression (A.P.) and Geometric Progression (G.P.) particularly. Both are extremely opposite to each other, say, its terminology of sum of sequence, etc. such as, the sequence of G.P. is assumed as $ \dfrac{a}{r},a,ar $ respectively. As a result, using factorization methods i.e. either by splitting/extracting the middle term or by using\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]so as to be sure of our final answer.