If $a, x, y, z, \mathrm{~b}$ are in A.P., then $x+y+z=12$ and if $a, x, y, z, b$ are in H.P, then $\

Question

If $a, x, y, z, \mathrm{~b}$ are in A.P., then $x+y+z=12$ and if $a, x, y, z, b$ are in H.P, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5}$. Then the value of $a^2+b^2=$

Professor PC Thomas Classes · Answer

$\frac{3}{2}(a+b)=12$ and $\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{4}{5}$