Question

If \[\dfrac{{(x + y)}}{2}\] , \[y\], \[\dfrac{{(y + z)}}{2}\] are in $HP$ , then \[x\],\[y\],\[z\] are in
A. AP
B. GP
C. HP
D. None of these

Answer 1

Hint: In this question, we are given some terms in HP, we have to find whether the terms \[x\],\[y\],\[z\]are 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 \[x\],\[y\],\[z\]. 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 \[\dfrac{{(x + y)}}{2}\] , \[y\], \[\dfrac{{(y + z)}}{2}\]. We know that reciprocal terms in HP are in AP.
Hence , the terms \[\dfrac{2}{{(x + y)}}\] , \[\dfrac{1}{y}\], \[\dfrac{2}{{(y + z)}}\] are in AP.
We also know that the difference of terms is constant in AP.
Hence, \[ \Rightarrow \;\dfrac{1}{y} - \dfrac{2}{{(x + y)}} = \dfrac{2}{{(y + z)}} - \dfrac{1}{y}\]
Taking LCM of denominator we have ,
\[ \Rightarrow \;\dfrac{{(x + y) - 2y}}{{y(x + y)}} = \dfrac{{2y - (y + z)}}{{y(y + z)}}\]
Cancelling \[y\]in denominator both side and simplifying we have ,
\[ \Rightarrow \;\dfrac{{(x - y)}}{{(x + y)}} = \dfrac{{(y - z)}}{{(y + z)}}\]

Cross multiplying we get,
\[ \Rightarrow (x - y)(y + z) = (x + y)(y - z)\]
On solving we have
\[ \Rightarrow \;xy - {y^2} + xz - yz = xy + {y^2} - xz - yz\]
Cancelling \[xy\]and \[ - yz\] both side, we have
\[ \Rightarrow \; - {y^2} + xz = {y^2} - xz\]
\[ \Rightarrow \;2{y^2} = 2xz\]
On cancelling \[2\] both side , we get,
\[ \Rightarrow \;{y^2} = xz\]
This can also be written as
\[ \therefore \;\dfrac{y}{x} = \dfrac{z}{y}\]
This is the condition for terms in GP. (i.e. ratio of consecutive terms in GP are constant. )
Hence the terms \[x\],\[y\],\[z\]are in GP.

Hence, option B is correct.

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