Question

If $a,b,c$ be in GP and $a + x,b + x,c + x$, in HP, then the value of $x$ is ($a,b,c$ are distinct numbers)
1) $c$
2) $b$
3) $a$
4) None of these

Answer 1

Hint: We know the reciprocal of terms in HP form an AP. So, we will use this condition to find the arithmetic mean of the terms as we know the difference between the consecutive terms of an AP are equal. Also we have the condition that the terms $a,b,c$ be in GP. We can use this condition to get a relation between these three terms, which we can further replace in the arithmetic mean obtained. Using these conditions we can find the required value.

Complete step-by-step answer:
We are given that, $a,b,c$ be in GP.
Therefore, we can write, the geometric mean of the three numbers is,
\[{b^2} = ac - - - \left( 1 \right)\]
Also, given, $a + x,b + x,c + x$ are in HP.
We know the reciprocal of terms in HP form an AP.
Therefore, we can write, $\dfrac{1}{{a + x}},\dfrac{1}{{b + x}},\dfrac{1}{{c + x}}$ are in AP.
We know, the difference between the consecutive terms of an AP are equal.
Therefore, about this AP, we can write,
$\dfrac{1}{{b + x}} - \dfrac{1}{{a + x}} = \dfrac{1}{{c + x}} - \dfrac{1}{{b + x}}$
Taking LCM in both sides, we get,
$ \Rightarrow \dfrac{{\left( {a + x} \right) - \left( {b + x} \right)}}{{\left( {b + x} \right)\left( {a + x} \right)}} = \dfrac{{\left( {b + x} \right) - \left( {c + x} \right)}}{{\left( {c + x} \right)\left( {b + x} \right)}}$
Now, cancelling $\left( {b + x} \right)$ from the denominator, we get,
$ \Rightarrow \dfrac{{\left( {a + x} \right) - \left( {b + x} \right)}}{{\left( {a + x} \right)}} = \dfrac{{\left( {b + x} \right) - \left( {c + x} \right)}}{{\left( {c + x} \right)}}$
Now, opening the brackets, we get,
$ \Rightarrow \dfrac{{a + x - b - x}}{{a + x}} = \dfrac{{b + x - c - x}}{{c + x}}$
Further simplifying, we get,
$ \Rightarrow \dfrac{{a - b}}{{a + x}} = \dfrac{{b - c}}{{c + x}}$
Now, cross multiplying, we get,
$ \Rightarrow \left( {a - b} \right)\left( {c + x} \right) = \left( {a + x} \right)\left( {b - c} \right)$
Opening the brackets and simplifying, we get,
$ \Rightarrow ac + ax - bc - bx = ab - ac + bx - cx$
Now, taking all the terms on the left hand side, we get,
$ \Rightarrow 2ac + ax - bc - 2bx - ab + cx = 0$
Taking the coefficients of $x$ together, we get,
$ \Rightarrow 2ac - bc - ab + x\left( {a - 2b + c} \right) = 0$
Substituting $\left( 1 \right)$, we get,
$ \Rightarrow 2{b^2} - bc - ab + x\left( {a - 2b + c} \right) = 0$
Taking, $b$ common from first three terms we get,
$ \Rightarrow b(2b - c - a) + x\left( {a - 2b + c} \right) = 0$
Taking the $x$ term in right hand side, we get,
$ \Rightarrow b(2b - c - a) = - x\left( {a - 2b + c} \right)$
$ \Rightarrow b(2b - c - a) = x\left( {2b - c - a} \right)$
Cancelling the similar terms on both sides, we get,
$ \Rightarrow b = x$
$ \Rightarrow x = b$
Therefore, the correct option is 2.
So, the correct answer is “Option 2”.

Note: -Here we used the fact that the numbers in HP, their reciprocals are in AP, then took their differences as equal and solved. We can also directly use the harmonic mean of the three numbers and solve the problem. The formula of harmonic mean is,
$y = \dfrac{{2xz}}{{x + z}}$
Where, $x,y,z$ are in HP.