QUESTION

# If x, 1, z are in AP and x, 2, z are in GP, then x, 4, z will be inA. APB. GPC. HPD. None of these.

Hint: In order to solve this problem we need to apply the conditions of AP and GP when the three numbers of the series are given. Then we have to check whether the asked number is in AP, GP or HP.

As we know that if a, b, c are in AP then ${\text{2b = a + c}}$.
So, x, 1, z are in AP then 2(1) = x + z ………….(1)
We also know that if p, q, r are in GP then ${{\text{q}}^{\text{2}}}{\text{ = pr}}$
And it is given that x, 2, z are in GP the ${{\text{2}}^{\text{2}}}{\text{ = 4 = xz}}$………..(2)
So, we have to find about x, 4, z.
Let us check whether it is in AP or not.
As we know x + z = 2
But if x, 4, z is in AP then x + z = 4(2)=8
So, it is not in AP.
Now, let us check whether it is in GP or not.
As we know xz = 4
But if x, 4, z are in GP so xz = 16
So, it is not in the GP.
Now we will check it for HP.
So, $\dfrac{{\text{1}}}{{\text{x}}}{\text{,}}\,\dfrac{{\text{1}}}{{\text{4}}}{\text{,}}\,\dfrac{{\text{1}}}{{\text{y}}}$ will be in AP if x, 4, z is in HP.
So applying the condition of AP in $\dfrac{{\text{1}}}{{\text{x}}}{\text{,}}\,\dfrac{{\text{1}}}{{\text{4}}}{\text{,}}\,\dfrac{{\text{1}}}{{\text{y}}}$
$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{4} = \dfrac{1}{2}$…………(3)
Now on solving the LHS we get
$\Rightarrow \dfrac{{x + z}}{{xz}}$ which is equal to $\dfrac{2}{4} = \dfrac{1}{2}$ (From 1, 2) and From (3) we can say the value is same.
So, x. 4, z are in HP.

Note: To solve such problems we need to use the concept that if a, b, c are in AP the 2b = a + c and if a, b, c are in GP then ${{\text{b}}^{\text{2}}}{\text{ = ac}}$ and if a, b, c are in HP then $\dfrac{{\text{1}}}{{\text{a}}}{\text{,}}\,\dfrac{{\text{1}}}{{\text{b}}}{\text{,}}\,\dfrac{{\text{1}}}{{\text{c}}}$ are in AP.