The value of  $x.y.z=55or\frac{343}{55}$ according as the series $a,x,y,z,b$  form an A.P or H.P respectively, where a and b are positive natural numbers. Then the sum of  $a+b=$

If a,x, y,z, b to are in A.P then the common difference d of the AP is given by 
 $b=a+4d\Rightarrow d=\frac{b-a}{4}$

$\therefore x=a+d=\frac{a+b-a}{4}=\frac{b+3a}{4}$
 
 $y=a+2d=\frac{a+b-a}{2}=\frac{a+b}{2}$
 
 $z=a+3d=a+3\left(\frac{b-a}{4}\right)=\frac{a+3b}{4}$
$\therefore xyz=\frac{b+3a}{4}\times \frac{a+b}{2}\times \frac{a+3b}{4}$ 
 $\Rightarrow 55=\frac{(3a+b)(a+b)(a+3b)}{32}$

$\Rightarrow (3a+b)(a+b)(a+3b)=55\times 32$
When a,x,y,z,b are in H.P . Then 
$\frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{b}$  are in AP
Let D be the common different of this A.P . Then 
 $\frac{1}{b}=\frac{1}{a}+4D\Rightarrow D=\frac{a-b}{4ab}$
 $\therefore \frac{1}{x}=\frac{1}{a}+D=\frac{1}{a}+\frac{a-b}{4ab}=\frac{3b+a}{4ab}$
 
 $\frac{1}{y}=\frac{1}{a}+2D=\frac{1}{a}=\frac{a-b}{2ab}=\frac{a+b}{2ab}$
 
 $\frac{1}{z}=\frac{1}{a}+3D=\frac{1}{a}=\frac{3(a-b)}{4ab}=\frac{3a+b}{4ab}$
 
  $\therefore \frac{1}{x}.\frac{1}{y}.\frac{1}{z}=\frac{(3a+b)(a+b)(3a+b)}{32a^3{b}^3}$
 $\Rightarrow \frac{1}{xyz}=\frac{(3a+b)(a+b(a+3b))}{32a^3{b}^3}$
 $\Rightarrow \frac{55}{343}=\frac{55\times 32}{32a^3{b}^3}$
$\Rightarrow$${\left(ab\right)}^3=7^3$

$\Rightarrow ab=7$

$\Rightarrow a=a,b=7,ora=7,b=1$