Answer
Verified461.7k+ views
Hint: Geometric progression (GP) is a type of sequence, where each succeeding term is obtained by multiplying each preceding term by a fixed number, which is called a common ratio.
If three numbers $a$,$b$ and $c$ are in GP; then they must follow the relation: ${b^2} = ac$
Complete step-by-step answer:
Given; $k - 1$, $k + 2$ and $3k$ are in GP.
We know that if three numbers $a$,$b$ and $c$; then ${b^2} = ac$
On applying the above relation to the given terms $k - 1$, $k + 2$ and $3k$; we get-
${\left( {k + 2} \right)^2} = \left( {k - 1} \right)\left( {3k} \right)$
$ \Rightarrow {k^2} + 4k + 4 = 3{k^2} - 3k$
$ \Rightarrow {k^2} - 3{k^2} + 4k + 3k + 4 = 0$
$ \Rightarrow - 2{k^2} + 7k + 4 = 0$
On multiplying by $ - 1$, we get-
$ \Rightarrow 2{k^2} - 7k - 4 = 0$ ….. (1)
On using factorization method-
$ \Rightarrow 2{k^2} - \left( {8 - 1} \right)k - 4 = 0$
$ \Rightarrow 2{k^2} - 8k + k - 4 = 0$
$ \Rightarrow 2k\left( {k - 4} \right) + 1\left( {k - 4} \right) = 0$
$ \Rightarrow \left( {k - 4} \right)\left( {2k + 1} \right) = 0$
$ \Rightarrow k = 4$ or $k = \dfrac{{ - 1}}{2}$
Hence the value of $k$ will be $4$ or $\dfrac{{ - 1}}{2}$.
Note: We can also solve the above mentioned equation (1) by using the quadratic formula which is given by,
$x = $$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now compare the equation (1), i.e., $2{k^2} - 7k - 4 = 0$with the standard quadratic equation $a{x^2} + bx + c = 0$; we get-
$a = 2,b = - 7,c = - 4$ and $x = k$
On putting all the values, we get-
$k = \dfrac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4 \times 2 \times \left( { - 4} \right)} }}{{2 \times 2}}$
$ \Rightarrow $$k = \dfrac{{7 \pm \sqrt {49 + 32} }}{4}$
$ \Rightarrow $$k = \dfrac{{7 \pm \sqrt {81} }}{4}$
$ \Rightarrow $$k = \dfrac{{7 \pm 9}}{4}$
$ \Rightarrow $$k = \dfrac{{7 + 9}}{4}$ or $k = \dfrac{{7 - 9}}{4}$
$ \Rightarrow $$k = \dfrac{{16}}{4}$ or $k = \dfrac{{ - 2}}{4}$
$ \Rightarrow $$k = 4$ or $k = \dfrac{{ - 1}}{2}$
Hence the value of $k$ will be $4$ or $\dfrac{{ - 1}}{2}$.
If three numbers $a$,$b$ and $c$ are in GP; then they must follow the relation: ${b^2} = ac$
Complete step-by-step answer:
Given; $k - 1$, $k + 2$ and $3k$ are in GP.
We know that if three numbers $a$,$b$ and $c$; then ${b^2} = ac$
On applying the above relation to the given terms $k - 1$, $k + 2$ and $3k$; we get-
${\left( {k + 2} \right)^2} = \left( {k - 1} \right)\left( {3k} \right)$
$ \Rightarrow {k^2} + 4k + 4 = 3{k^2} - 3k$
$ \Rightarrow {k^2} - 3{k^2} + 4k + 3k + 4 = 0$
$ \Rightarrow - 2{k^2} + 7k + 4 = 0$
On multiplying by $ - 1$, we get-
$ \Rightarrow 2{k^2} - 7k - 4 = 0$ ….. (1)
On using factorization method-
$ \Rightarrow 2{k^2} - \left( {8 - 1} \right)k - 4 = 0$
$ \Rightarrow 2{k^2} - 8k + k - 4 = 0$
$ \Rightarrow 2k\left( {k - 4} \right) + 1\left( {k - 4} \right) = 0$
$ \Rightarrow \left( {k - 4} \right)\left( {2k + 1} \right) = 0$
$ \Rightarrow k = 4$ or $k = \dfrac{{ - 1}}{2}$
Hence the value of $k$ will be $4$ or $\dfrac{{ - 1}}{2}$.
Note: We can also solve the above mentioned equation (1) by using the quadratic formula which is given by,
$x = $$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now compare the equation (1), i.e., $2{k^2} - 7k - 4 = 0$with the standard quadratic equation $a{x^2} + bx + c = 0$; we get-
$a = 2,b = - 7,c = - 4$ and $x = k$
On putting all the values, we get-
$k = \dfrac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4 \times 2 \times \left( { - 4} \right)} }}{{2 \times 2}}$
$ \Rightarrow $$k = \dfrac{{7 \pm \sqrt {49 + 32} }}{4}$
$ \Rightarrow $$k = \dfrac{{7 \pm \sqrt {81} }}{4}$
$ \Rightarrow $$k = \dfrac{{7 \pm 9}}{4}$
$ \Rightarrow $$k = \dfrac{{7 + 9}}{4}$ or $k = \dfrac{{7 - 9}}{4}$
$ \Rightarrow $$k = \dfrac{{16}}{4}$ or $k = \dfrac{{ - 2}}{4}$
$ \Rightarrow $$k = 4$ or $k = \dfrac{{ - 1}}{2}$
Hence the value of $k$ will be $4$ or $\dfrac{{ - 1}}{2}$.