If a,;b,;c are in A.P., then {10^{ax + 10}},; | vedclass

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(c) $a,\;b,\;c$are in $A.P.$

$ \Rightarrow $$2b = a + c$

Now ${({10^{bx + 10}})^2} = ({10^{ax + 10}}.\;{10^{cx + 10}})$

$ \Rightarrow $ ${10^

Vedclass · Accepted Answer

$G.P.$ for all values of $x$

Vedclass · Answer

(c) $a,\;b,\;c$are in $A.P.$

$ \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 $r$.

If $a,\;b,\;c$ are in $A.P.$, then ${10^{ax + 10}},\;{10^{bx + 10}},\;{10^{cx + 10}}$ will be in

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