Mathematics - Sequence & Series Question with Solution | TestHub

For (P), if $a,b,c$ are in A.P., then $2b=a+c$. So $a-2b+c=a-(a+c)+c=0$. So (P)-(3).

For (Q), if $x,y,z$ are in G.P., then $y^2=xz$. Taking logarithm on both sides, $2\log y=\log x+\log z$, which is the condition for $\log x,\log y,\log z$ to be in A.P. So (Q)-(1). For (R), by AM-GM inequality, $\frac{a+b+c}{3}\ge \sqrt[3]{abc}$. Given $a+b+c=12$, so $\frac{12}{3}\ge \sqrt[3]{abc}\Rightarrow 4\ge \sqrt[3]{abc}\Rightarrow 64\ge abc$. The maximum value of $abc$ is 64. So (R)-(2).

For (S), if $a,b,c$ are in H.P., consider $a=2,b=3,c=6$. Then $a^2=4,b^2=9,c^2=36$. These are not in A.P. ($9-4=5,36-9=27$), not in G.P. ($9/4\ne 36/9$), and not in H.P. ($1/4,1/9,1/36$ are not in A.P.). So (S)-(4).