Question

How do you find $k$ such that $k+1,4k,3k+5$ are in geometric sequence?

Answer 1

Hint: To solve this question we need to know about the Geometric Progression. To solve the question, the concept of common ratio should be known. Common ratio is the fraction having a numerator as ${{(n+1)}^{th}}$ term and the denominator is the ${{(n)}^{th}}$. To solve the value of $k$ we need to solve the quadratic equation, which occurs due to cross multiplication.

Complete step-by-step solution:
We are asked to find the value of $k$, for the given question. We will need to equate the common ratio. Since the three numbers given in terms of $k$ are in Geometric sequence the ratio of the two numbers will be equal. Hence mathematically it could be written as,
$\dfrac{4k}{k+1}=\dfrac{3k+5}{4k}$
First step to find the value of $k$ is to perform cross multiplication
$4k\times 4k=(3k+5)(k+1)$
On multiplying each terms in the equation we get,
$\Rightarrow 16{{k}^{2}}=3{{k}^{2}}+8k+5$
Further,
$\Rightarrow 16{{k}^{2}}-3{{k}^{2}}-8k-5=0$
$\Rightarrow 13{{k}^{2}}-8k-5=0$
To find the value of $k$ we will have know the roots, for which we need to factorise the given polynomial $13{{k}^{2}}-8k-5$
$13{{k}^{2}}-8k-5$
On middle term factoring the polynomial we get,
$\Rightarrow 13{{k}^{2}}-13k+5k-5$
Taking $13$ and $5$ common we get the following expression:
$\Rightarrow 13k(k-1)+5(k-1)$
$\Rightarrow (13k+5)(k-1)$
So the roots we get are $k=\dfrac{-5}{13}$ and$k=1$.
$\therefore $ The value of $k$ for which the three numbers are in geometric sequence are $k=\dfrac{-5}{13}$ and $k=1$.

Note: To check whether the value of $k$ is correct or not we need to put the values of $k$ in the given three numbers. The numbers are $k+1,4k,3k+5$.
$\dfrac{4k}{k+1}=\dfrac{3k+5}{4k}$
Put the value of $k$ on the given equation:
$\Rightarrow \dfrac{4\times 1}{1\times 1}=\dfrac{3\times 1+5}{4\times 1}$
On calculating we get ,
$\Rightarrow \dfrac{4}{2}=\dfrac{8}{4}$
$\Rightarrow 2=2$
$\therefore $ The two ratios are same for $k=1$
Similarly for $k=\dfrac{-5}{13}$
$\Rightarrow \dfrac{4\times \dfrac{-5}{13}}{\dfrac{-5}{13}+1}=\dfrac{3\times \dfrac{-5}{13}+5}{4\times \dfrac{-5}{13}}$
On calculating further,
$\Rightarrow \dfrac{\dfrac{-20}{13}}{\dfrac{-5+13}{13}}=\dfrac{\dfrac{-15+65}{13}}{\dfrac{-20}{13}}$
$\Rightarrow \dfrac{-20}{8}=\dfrac{50}{-20}$
$\Rightarrow \dfrac{-5}{2}=\dfrac{-5}{2}$
$\therefore $ The ratio is the same for $k=\dfrac{-5}{13}$, the value of $k$ is the same.
By solving this question we need to keep in mind that the ratio of ${{(N+1)}^{th}}$ to ${{N}^{th}}$ should be equal to ${{N}^{th}}$ to ${{(N-1)}^{th}}$and not the vice-versa.