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Question: If \(x,2y,3z\) are in AP, where the distinct numbers \(x,y,z\) are in GP, then the common ratio is: ...

If x,2y,3zx,2y,3z are in AP, where the distinct numbers x,y,zx,y,z are in GP, then the common ratio is:
A. 33
B. 13\dfrac{1}{3}
C. 22
D. 12\dfrac{1}{2}

Explanation

Solution

We are given that certain series are in AP and some in GP. Writing the formulas for the AP and GP. Which will form two equations and by using the formula for common ratio which will form a quadratic equation. By solving the quadratic equation, we can find or answer.

Complete step by step solution:
Firstly, learning the meaning of AP and GP. AP full form is arithmetic progression. AP means sequence of numbers having the same difference between the consecutive numbers. It can be written as a,a+d,a+2d,…,…a,a + d,a + 2d, \ldots,\ldots where d is the common difference.
It can be written as 2(a+d)=a+(a+2d)2\left( {a + d} \right) = a + \left( {a + 2d} \right)
Whereas GP full form is geometric progression which means if in a sequence of terms, each succeeding term is generated by multiplying each preceding term with the constant value and that constant value is known as the common ratio. It can be written in the form of a,ar,ar2,ar3……a,ar,a{r^2},a{r^3} \ldots \ldots where r is the common ratio.
It can be written as (ar)2=(a)(ar2){\left( {ar} \right)^2} = \left( a \right)\left( {a{r^2}} \right)
According to the question, we are provided that x,2y,3zx,2y,3zare in AP. So, these numbers can be written as
2(2y)=x+3z2\left( {2y} \right) = x + 3z …(1) \ldots \left( 1 \right)
And x,y,zx,y,zare in GP. So, these can be written as y2=xz{y^2} = xz …(2) \ldots \left( 2 \right)
In the question, we have to find the common ratio. So, let the common ratio be rr and it should be equal to the r=yxr = \dfrac{y}{x} and r2=zx{r^2} = \dfrac{z}{x}
In order to find r, dividing the equation (1)\left( 1 \right) by x
4yx=1+3zx4\dfrac{y}{x} = 1 + 3\dfrac{z}{x} and now putting the value of r,
4r=1+3r24r = 1 + 3{r^2} or 3r2−4r+1=03{r^2} - 4r + 1 = 0
Solving the above quadratic equation,
3r2−3r−r+1=0 3r(r−1)−1(r−1)=0 (3r−1)(r−1)=0  3{r^2} - 3r - r + 1 = 0 \\\ 3r\left( {r - 1} \right) - 1\left( {r - 1} \right) = 0 \\\ \left( {3r - 1} \right)\left( {r - 1} \right) = 0 \\\
Hence, r=1,13r = 1,\dfrac{1}{3}
But r=1r = 1 is neglected as x,y,zx,y,z are distinct.
Hence, the common ratio is 13\dfrac{1}{3}. So, the correct option is B.

Note: Be careful while rejecting any value as in the above question, we get two values for the common ratio. But we had rejected one value due to the values becoming the same by putting the value into the question. The examiner can confuse the students by giving that value into the options.