The value of $x+y+z$ is 15 . If $a, x, y, z, b$ are in A.P. while the value of $\frac{1}{x}+\frac{1}

Question

The value of $x+y+z$ is 15 . If $a, x, y, z, b$ are in A.P. while the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $\frac{5}{3}$. If $a, x, y, z, b$ are in H.P. then  $a+b$ =? .

Professor PC Thomas Classes · Answer

(i) $a, x, y, z, b$ are in A.P. Let $d$ be the common difference\ $a+b=(a+x+y+z+b)-(x+y+z)$\ $=(a+a+d+a+2 d+a+3 d+a+4 d)-15$\ $=5 a+10 d-15$\ But $a+4 d=b \Rightarrow d=\frac{b-a}{4}$\ $	herefore a+b=5 a+10\left(\frac{b-a}{4}ight)-15=\frac{5}{2}(a+b)-15$ $$\begin{equation*}\Rightarrow a+b=10 	ag{i} \end{equation*}$$ $a, x, y, z, b$ are in H.P.\ $\Rightarrow \frac{1}{a}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{b}$ are in A.P.\ As before $\frac{1}{a}+\frac{1}{b}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}ight)-\frac{5}{3}$ $$\begin{equation*}\Rightarrow \frac{a+b}{a b}=\frac{10}{9} \quad 	herefore a b=9 	ag{ii} \end{equation*}$$ Solving (i) and (ii), $a=1, b=9$ or $a=9, b=1$\ (ii) $	herefore x, y, z$ are in H.P. So, $\quad y=\frac{2 x z}{x+z}$\ L.H.S. $=\log (x+z)+\log (x+z-2 y)$ $$ =\log (x+z)+\log \left(x+z-\frac{4 x z}{x+z}ight) $$ $=\log (x+z)+\log \left(\frac{x^{2}+z^{2}+2 x z-4 x z}{x+z}ight)$\ $=\log (x+z)+\log (x-z)^{2}-\log (x+z)$\ $=\log (x-z)^{2}=2 \log (x-z)=$ R.H.S.\