Question

Let a, x, b be in A.P; a, y, b be in G.P and a, z, b be in H.P. If x = y + 2 and a = 5z then:
\[\left( a \right){y^2} = xz\]
$\left( b \right)x > y > z$
$\left( c \right)a = 9,b = 1$
$\left( d \right)a = \dfrac{9}{4},b = \dfrac{1}{4}$

Answer 1

Hint: In this particular type of question use the concept that in A.P common difference is equal, in G.P common ratio is equal and H.P is the reciprocal of the A.P so use this properties construct the equation and try to simplify these equation using the given condition so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given
a, x, b......................... A.P
Therefore, according to the property of Arithmetic progression common difference is equal
Therefore,
x – a = b – x
$ \Rightarrow 2x = a + b$................ (1)
Now it is also given that
a, y, b......................... G.P
Therefore, according to the property of Geometric progression common ratio is equal
$ \Rightarrow \dfrac{y}{a} = \dfrac{b}{y}$
$ \Rightarrow {y^2} = ab$................. (2)
Now it is also given that
a, z, b be in H.P
Now as we know that the harmonic progression is the reciprocal of arithmetic progression.
$ \Rightarrow \dfrac{2}{z} = \dfrac{1}{a} + \dfrac{1}{b}$................... (3)
Now it is given that
x = y + 2 and a = 5z
Now substitute this value in equation (1), (2) and (3) we have,
$ \Rightarrow 2\left( {y + 2} \right) = 5z + b$............. (4),
$ \Rightarrow {y^2} = 5zb$.............. (5)
And
$ \Rightarrow \dfrac{2}{z} = \dfrac{1}{{5z}} + \dfrac{1}{b}$
$ \Rightarrow \dfrac{2}{z} = \dfrac{{5z + b}}{{5zb}}$
$ \Rightarrow 5b = \dfrac{{5z + b}}{2}$..................... (6)
Now from equation (4) we have,
$ \Rightarrow y = \dfrac{{5z + b}}{2} - 2$
$ \Rightarrow y = 5b - 2$.............. (7)
Now substitute this value in equation (5) we have,
$ \Rightarrow {\left( {5b - 2} \right)^2} = 5zb$
Now simplify this we have,
$ \Rightarrow 25{b^2} + 4 - 20b = 5zb$
$ \Rightarrow z = 5b + \dfrac{4}{5}b - 4$............. (8)
Now substitute this value in equation (6) we have,
$ \Rightarrow 5b = \dfrac{{5\left( {5b + \dfrac{4}{5}b - 4} \right) + b}}{2}$
$ \Rightarrow 5b = \dfrac{{25b + 4b - 20 + b}}{2}$
$ \Rightarrow 10b = 30b - 20$
$ \Rightarrow 20b = 20$
$ \Rightarrow b = 1$
Now from equation (8) we have,
$ \Rightarrow z = 5 + \dfrac{4}{5} - 4 = \dfrac{{25 + 4 - 20}}{5} = \dfrac{9}{5}$
Now from equation (7) we have,
$ \Rightarrow y = 5 - 2 = 3$
Now from equation (2) we have,
$ \Rightarrow {3^2} = a\left( 1 \right)$
$ \Rightarrow a = 9$
Now from equation (1) we have,
$ \Rightarrow 2x = 9 + 1$
$ \Rightarrow x = \dfrac{{10}}{2} = 5$
So the values is given as
a = 9, b = 1, x = 5, y = 3 and z = (9/5)
So, option \[\left( a \right){y^2} = xz\]
$ \Rightarrow {3^2} = 5\left( {\dfrac{9}{5}} \right)$
$ \Rightarrow 9 = 9$
So, option (a) is correct.
Now, option b, x > y > z
So as from the calculated values we see that this option is also correct.
And option (c) is also correct.
But option (d) is not correct.
Hence options (a), (b) and (c) are the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the properties of A.P, G.P and H.P which is stated above then first write the equations using these properties as above then substitute the given condition in these equation and construct the equation which has only one variable as above calculated then simplify we will get all the values of the variables then check the options one by one we will get the required answer.