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Question: If \(k - 1\), \(k + 2\) and \(3k\) are in GP, find the value of \(k\)....

If k1k - 1, k+2k + 2 and 3k3k are in GP, find the value of kk.

Explanation

Solution

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 aa,bb and cc are in GP; then they must follow the relation: b2=ac{b^2} = ac

Complete step-by-step answer:
Given; k1k - 1, k+2k + 2 and 3k3k are in GP.
We know that if three numbers aa,bb and cc; then b2=ac{b^2} = ac
On applying the above relation to the given terms k1k - 1, k+2k + 2 and 3k3k; we get-
(k+2)2=(k1)(3k){\left( {k + 2} \right)^2} = \left( {k - 1} \right)\left( {3k} \right)
k2+4k+4=3k23k\Rightarrow {k^2} + 4k + 4 = 3{k^2} - 3k
k23k2+4k+3k+4=0\Rightarrow {k^2} - 3{k^2} + 4k + 3k + 4 = 0
2k2+7k+4=0\Rightarrow - 2{k^2} + 7k + 4 = 0
On multiplying by 1 - 1, we get-
2k27k4=0\Rightarrow 2{k^2} - 7k - 4 = 0 ….. (1)
On using factorization method-
2k2(81)k4=0\Rightarrow 2{k^2} - \left( {8 - 1} \right)k - 4 = 0
2k28k+k4=0\Rightarrow 2{k^2} - 8k + k - 4 = 0
2k(k4)+1(k4)=0\Rightarrow 2k\left( {k - 4} \right) + 1\left( {k - 4} \right) = 0
(k4)(2k+1)=0\Rightarrow \left( {k - 4} \right)\left( {2k + 1} \right) = 0
k=4\Rightarrow k = 4 or k=12k = \dfrac{{ - 1}}{2}

Hence the value of kk will be 44 or 12\dfrac{{ - 1}}{2}.

Note: We can also solve the above mentioned equation (1) by using the quadratic formula which is given by,
x=x = b±b24ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Now compare the equation (1), i.e., 2k27k4=02{k^2} - 7k - 4 = 0with the standard quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0; we get-
a=2,b=7,c=4a = 2,b = - 7,c = - 4 and x=kx = k
On putting all the values, we get-
k=(7)±(7)24×2×(4)2×2k = \dfrac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4 \times 2 \times \left( { - 4} \right)} }}{{2 \times 2}}
\Rightarrow k=7±49+324k = \dfrac{{7 \pm \sqrt {49 + 32} }}{4}
\Rightarrow k=7±814k = \dfrac{{7 \pm \sqrt {81} }}{4}
\Rightarrow k=7±94k = \dfrac{{7 \pm 9}}{4}
\Rightarrow k=7+94k = \dfrac{{7 + 9}}{4} or k=794k = \dfrac{{7 - 9}}{4}
\Rightarrow k=164k = \dfrac{{16}}{4} or k=24k = \dfrac{{ - 2}}{4}
\Rightarrow k=4k = 4 or k=12k = \dfrac{{ - 1}}{2}
Hence the value of kk will be 44 or 12\dfrac{{ - 1}}{2}.