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NCERT Exemplar · Q20

Q.If x,2y,3zx, 2y, 3z are in A.P., where the distinct numbers x,y,zx, y, z are in G.P. then the common ratio of the G.P. is
(A) 33
(B) 13\dfrac{1}{3}
(C) 22
(D) 12\dfrac{1}{2}

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The problem gives two conditions: x,2y,3zx, 2y, 3z are in arithmetic progression, and x,y,zx, y, z (distinct) are in geometric progression. Using the AP property 2⋅(2y)=x+3z2 \cdot (2y) = x + 3z and the GP property y2=xzy^2 = xz, we solve for the common ratio r=y/xr = y/x. The valid distinct ratio is 13\boxed{\frac{1}{3}}, which corresponds to option (B).

We have two sequences here, and the trick is to connect them through the common terms x,y,zx, y, z. Let’s start with what each progression tells us.

Geometric Progression (GP): If x,y,zx, y, z are in GP, then the ratio between consecutive terms is constant. Let that common ratio be rr. So:

y=xr,z=xr2y = xr, \quad z = xr^2

This is the cleanest way to express all three in terms of xx and rr. Since x,y,zx, y, z are distinct, r≠1r \neq 1 and x≠0x \neq 0 (otherwise all would be zero, not distinct).

Arithmetic Progression (AP): The terms x,2y,3zx, 2y, 3z are in AP. For three numbers a,b,ca, b, c to be in AP, the middle term is the average of the other two:

2b=a+c2b = a + c

Here a=xa = x, b=2yb = 2y, c=3zc = 3z. So:

2⋅(2y)=x+3z⇒4y=x+3z2 \cdot (2y) = x + 3z \quad \Rightarrow \quad 4y = x + 3z

Now substitute the GP expressions into this AP equation.

  1. Substitute y=xry = xr and z=xr2z = xr^2 into 4y=x+3z4y = x + 3z:

4(xr)=x+3(xr2)4(xr) = x + 3(xr^2)

Since x≠0x \neq 0, divide through by xx:

4r=1+3r24r = 1 + 3r^2

  1. Rearrange into a quadratic equation:

3r2−4r+1=03r^2 - 4r + 1 = 0

  1. Solve the quadratic:

r=4±16−126=4±26r = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6}

So:

r=4+26=1orr=4−26=13…r = \frac{4 + 2}{6} = 1 \quad \text{or} \quad r = \frac{4 - 2}{6} = \frac{1}{3} …

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