Question

If k-2, 2k + 1 and 6k +3 are in G.P. find k
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Answer 1

Answer:

${(2k + 1)}^{2} = (k - 2)(6k + 3) \\ \\ 4 {k}^{2} + 1 + 4k = 6 {k}^{2} - 9k - 6 \\ \\ - 2 { k}^{2} + 13k + 7 = 0 \\ \\ 2 {k}^{2} - 13k - 7 = 0 \\ \\ 2 {k}^{2} - 14k + k - 7 = 0 \\ \\ 2k(k - 7) + 1(k - 7) = 0 \\ \\ (2k + 1)(k - 7) = 0 \\ \\ k = 7 \: \: \: \: \: \: \: \: k = - \frac{1}{2}$

Answer 2

The value of k is 7 or -0.5.

Step-by-step explanation:

Given : If  k-2, 2k + 1 and 6k +3 are in G.P.

To find : The value of k ?

Solution :

If a,b,c are in G.P then $b^2=ac$.

On comparing, $a=k-2, b=2k + 1,c=6k +3$

Substitute in the formula,

$(2k+1)^2=(k-2)(6k+3)\\\\4k^2+1+4k=6k^2+3k-12k-6\\\\6k^2-4k^2+3k-12k-4k-6-1=0\\\\2k^2-13k-7=0$

Applying quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=2, b=-13 and c=-7

$x=\frac{-(-13)\pm\sqrt{(-13)^2-4(2)(-7)}}{2(2)}\\\\x=\frac{13\pm\sqrt{225}}{4}\\\\x=\frac{13\pm15}{4}\\\\x=\frac{13+15}{4},\frac{13-15}{4}\\\\x=7,-0.5$

Therefore, the value of k is 7 or -0.5.

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