Question

The value of x+y+z=15, if a,x,y,z,b are in A.P, while the value of $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{3}$ if $a,x,y,z,b$ are in H.P. Find $a\times b$.

Answer 1

Hint: For the above question we will use the concept that if ‘a’ and ‘b’ are two terms of an A.P and there are ‘n’ arithmetic means say ${{x}_{1}},{{x}_{2}},{{x}_{3}},......{{x}_{n}}$ between them then their sum is equal to n time the average value of ‘a’ and ‘b’.
$\Rightarrow {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{n}}=n\left( \dfrac{a+b}{2} \right)$
Also, we know that if ${{a}_{1}},{{a}_{2}},{{a}_{3}},......$ are in H.P in $\dfrac{1}{{{a}_{1}}},\dfrac{1}{{{a}_{2}}},\dfrac{1}{{{a}_{3}}},.....$ must be in A.P, we will also use this concept.

Complete step-by-step solution -
We have been given that if \[x+y+z=15\], if $a,x,y,z,b$ are in A.P, while the value of $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{3}$ if $a,x,y,z,b$ are in H.P.
We know that if a and b are in A.P and there are n arithmetic mean say ${{x}_{1}},{{x}_{2}},{{x}_{3}},......{{x}_{n}}$ between them then,
$\Rightarrow {{x}_{1}}+{{x}_{2}}+{{x}_{3}}........... +{{x}_{n}}=n\left( \dfrac{a+b}{2} \right)$
We have 3 arithmetic mean between a and b which are $x,y,z$.
$\begin{align}
  & \Rightarrow x+y+z=3\left( \dfrac{a+b}{2} \right) \\
 & \Rightarrow 15=3\left( \dfrac{a+b}{2} \right) \\
 & \Rightarrow \dfrac{15\times 2}{3}=a+b \\
 & \Rightarrow 10=a+b \\
 & \Rightarrow a+b=10...........\left( 1 \right) \\
\end{align}$
We have $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{3}$ if $a,x,y,z,b$ are in H.P.
We know that if ${{a}_{1}},{{a}_{2}},{{a}_{3}},......{{a}_{n}}$ are in H.P then their reciprocal i.e., $\dfrac{1}{{{a}_{1}}},\dfrac{1}{{{a}_{2}}},\dfrac{1}{{{a}_{3}}},.....\dfrac{1}{{{a}_{n}}}$ must be in A.P.
$\Rightarrow \dfrac{1}{a},\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z},\dfrac{1}{b}$ are in A.P.
We know that the sum of ‘n’ arithmetic mean between a and b of an A.P is given by,
 $\begin{align}
  & {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{n}}=n\left( \dfrac{a+b}{2} \right) \\
 & \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\left( \dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{2} \right) \\
 & \Rightarrow \dfrac{5}{3}=3\left( \dfrac{a+b}{2ab} \right) \\
 & \Rightarrow \dfrac{5\times 2}{3\times 3}=\dfrac{a+b}{ab} \\
\end{align}$
Using equation (1) we have $\left( a+b \right)=10$.
$\begin{align}
  & \Rightarrow \dfrac{10}{9}=\dfrac{10}{ab} \\
 & \Rightarrow ab=9 \\
\end{align}$
Therefore, the required value of $\left( a\times b \right)$is equal to 9.

Note: Remember the point that the reciprocal of the term of H.P are in A.P. If we have $x,y,z$ are in H.P then $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ must be in A.P. Also, remember that Arithmetic Progression is a sequence of numbers that is constant and it is denoted by (A.P) in short form whereas H.P is Harmonic Progression.