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Question: If \[\dfrac{{(x + y)}}{2}\] , \[y\], \[\dfrac{{(y + z)}}{2}\] are in \(HP\) , then \[x\],\[y\],\[z\]...

If (x+y)2\dfrac{{(x + y)}}{2} , yy, (y+z)2\dfrac{{(y + z)}}{2} are in HPHP , then xx,yy,zz are in
A. AP
B. GP
C. HP
D. None of these

Explanation

Solution

In this question, we are given some terms in HP, we have to find whether the terms xx,yy,zzare in either AP, GP, HP or None. We will proceed with given terms in HP , and write them in AP by taking reciprocal . Now we will use the condition for terms in AP (i.e. difference of two consecutive terms is constant) and solve to get a result in terms containing xx,yy,zz. Then we will check the result satisfying condition for either AP/GP/HP/NONE to get the final answer.

Complete step by step answer:
Consider the given question, the terms in HP are (x+y)2\dfrac{{(x + y)}}{2} , yy, (y+z)2\dfrac{{(y + z)}}{2}. We know that reciprocal terms in HP are in AP.
Hence , the terms 2(x+y)\dfrac{2}{{(x + y)}} , 1y\dfrac{1}{y}, 2(y+z)\dfrac{2}{{(y + z)}} are in AP.
We also know that the difference of terms is constant in AP.
Hence, ⇒  1y−2(x+y)=2(y+z)−1y \Rightarrow \;\dfrac{1}{y} - \dfrac{2}{{(x + y)}} = \dfrac{2}{{(y + z)}} - \dfrac{1}{y}
Taking LCM of denominator we have ,
⇒  (x+y)−2yy(x+y)=2y−(y+z)y(y+z)\Rightarrow \;\dfrac{{(x + y) - 2y}}{{y(x + y)}} = \dfrac{{2y - (y + z)}}{{y(y + z)}}
Cancelling yyin denominator both side and simplifying we have ,
⇒  (x−y)(x+y)=(y−z)(y+z)\Rightarrow \;\dfrac{{(x - y)}}{{(x + y)}} = \dfrac{{(y - z)}}{{(y + z)}}

Cross multiplying we get,
⇒(x−y)(y+z)=(x+y)(y−z)\Rightarrow (x - y)(y + z) = (x + y)(y - z)
On solving we have
⇒  xy−y2+xz−yz=xy+y2−xz−yz\Rightarrow \;xy - {y^2} + xz - yz = xy + {y^2} - xz - yz
Cancelling xyxyand −yz - yz both side, we have
⇒  −y2+xz=y2−xz\Rightarrow \; - {y^2} + xz = {y^2} - xz
⇒  2y2=2xz\Rightarrow \;2{y^2} = 2xz
On cancelling 22 both side , we get,
⇒  y2=xz\Rightarrow \;{y^2} = xz
This can also be written as
∴  yx=zy\therefore \;\dfrac{y}{x} = \dfrac{z}{y}
This is the condition for terms in GP. (i.e. ratio of consecutive terms in GP are constant. )
Hence the terms xx,yy,zzare in GP.

Hence, option B is correct.

Note: If the terms a,b,ca,b,c are in GP, Then their common ratio is constant . i.e. ba=cb\dfrac{b}{a} = \dfrac{c}{b}. If the terms a,b,ca,b,c are in AP , then their common difference is constant. i.e. b−a=c−bb - a = c - b. If the terms a,b,ca,b,c are in AP , then the reciprocal of terms are in GP. i.e. 1a,1b,1c\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} are in GP .