Mathematics - Sequence & Series Question with Solution | TestHub

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

Therefore, $x=a+d=\frac{a+\frac{b-a}{4}}{1}=\frac{4a+b-a}{4}=\frac{3a+b}{4}$.

$y=a+2d=a+2\left(\frac{b-a}{4}\right)=\frac{a+b}{2}$.

$z=a+3d=a+3\left(\frac{b-a}{4}\right)=\frac{4a+3b-3a}{4}=\frac{a+3b}{4}$.

Thus, $xyz=\frac{3a+b}{4}\times \frac{a+b}{2}\times \frac{a+3b}{4}$.

$\Rightarrow 55=\frac{(3a+b)(a+b)(a+3b)}{32}$

$\Rightarrow (3a+b)(a+b)(a+3b)=55\times 32$

When $a,x,y,z,b$ are in H.P., then $\frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{b}$ are in A.P. Let $D$ be the common difference of this A.P. Then $\frac{1}{b}=\frac{1}{a}+4D\Rightarrow D=\frac{a-b}{4ab}$.

Therefore, $\frac{1}{x}=\frac{1}{a}+D=\frac{1}{a}+\frac{a-b}{4ab}=\frac{4b+a-b}{4ab}=\frac{3b+a}{4ab}$.

$\frac{1}{y}=\frac{1}{a}+2D=\frac{1}{a}+2\left(\frac{a-b}{4ab}\right)=\frac{1}{a}+\frac{a-b}{2ab}=\frac{2b+a-b}{2ab}=\frac{a+b}{2ab}$.

$\frac{1}{z}=\frac{1}{a}+3D=\frac{1}{a}+3\left(\frac{a-b}{4ab}\right)=\frac{4b+3a-3b}{4ab}=\frac{3a+b}{4ab}$.

Thus, $\frac{1}{x}\cdot \frac{1}{y}\cdot \frac{1}{z}=\frac{(3b+a)(a+b)(3a+b)}{32a^3{b}^3}$.

$$\Rightarrow \frac{1}{xyz}=\frac{(3a+b)(a+b)(a+3b)}{32a^3{b}^3}$$

We are given that $xyz=\frac{343}{55}$ for the H.P. case.

$$\Rightarrow \frac{55}{343}=\frac{55\times 32}{32a^3{b}^3}$$

$$\Rightarrow (ab{)}^3=7^3$$

$$\Rightarrow ab=7$$

This implies $a=1,b=7$ or $a=7,b=1$.