Three numbers a,b and c are in geometric progression. If $4a,5band4c$  are in arithmetic progression and $a+b+c=70$ , then the value of $|c-a|$  is equal to  

$\begin{array}{l}a+b+c=70\Rightarrow a+ar+a{r}^2=70\\ Also,10b=4a+4c\Rightarrow 10ar=4a+4a{r}^2\\ \Rightarrow 2r^2-5r+2=0\Rightarrow r=2,\frac{1}{2}\\ For,r=2wehave,a(1+r+r^2)=70\Rightarrow a=\frac{70}{1+r+r^2}\\ =\frac{70}{1+2+4}=10\Rightarrow a=10,b=20,c=40\Rightarrow |c-a|=30\\ For,r=\frac{1}{2},wehavea=\frac{70}{1+r+r^2}=\frac{70}{1+\frac{1}{2}+\frac{1}{4}}=\frac{70\times 4}{4+2+1}=40\\ \Rightarrow b=20,c=10\Rightarrow |c-a|=|10-40|=30\\ So,inboththecases|c-a|=30\end{array}$