MCQ
If $a,\;b,\;c$ are in $A.P.$, then ${10^{ax + 10}},\;{10^{bx + 10}},\;{10^{cx + 10}}$ will be in
- A$A.P.$
- B$G.P.$ only when $x > 0$
- ✓$G.P.$ for all values of $x$
- D$G.P.$ for $x < 0$
$ \Rightarrow $$2b = a + c$
Now ${({10^{bx + 10}})^2} = ({10^{ax + 10}}.\;{10^{cx + 10}})$
$ \Rightarrow $ ${10^{2(bx + 10)}} = {10^{ax + cx + 20}}$
$ \Rightarrow $ $2(bx + 10) = ax + cx + 20,\;\rlap{--} V\,x$
$ \Rightarrow $ $2b = a + c\;\;i.e.\;\;a,\;b,\;c$ are in $A.P.$
Hence these are in $G.P.$ $\forall x$.
Note : As we know if $a,\;b,\;c$ are in $A.P.$,
then ${x^{an + r}},\;{x^{bn + r}},\;{x^{cn + r}}$ are in $G.P. $ for every $n$ and $r$.
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