Mathematics - Sequence & Series Question with Solution | TestHub

Given $a,b,c$ are in A.P., so $2b=a+c$ (1). Given $b,c,d$ are in G.P., so $c^2=bd$ (2). Given $c,d,e$ are in H.P., so $\frac{2}{d}=\frac{1}{c}+\frac{1}{e}=\frac{e+c}{ce}$, which implies $d=\frac{2ce}{c+e}$ (3). Substitute $d$ from (3) into (2):$$c^2=b\left(\frac{2ce}{c+e}\right)$$

Assuming $c\ne 0$ (if $c=0$, then $b=0,d=0,a=0,e=0$, which is a trivial case where all numbers are zero and in all progressions):$$c=b\left(\frac{2e}{c+e}\right)$$ $$c(c+e)=2be⟹c^2+ce=2be$$ (4) Substitute $b$ from (1) into (4):$$c^2+ce=2\left(\frac{a+c}{2}\right)e$$ $$c^2+ce=(a+c)e$$ $$c^2+ce=ae+ce$$ $$c^2=ae$$ The condition $c^2=ae$ implies that $a,c,e$ are in Geometric Progression.