A) A.P.
B) G.P. only when \[x>0\]
C) G.P. for all values of \[x\]
D) G.P. for \[x<0\]
Correct Answer: C
Solution :
\[a,\ b,\ c\]are in A.P. \[\Rightarrow \]\[2b=a+c\] Now \[{{({{10}^{bx+10}})}^{2}}=({{10}^{ax+10}}.\ {{10}^{cx+10}})\] \[\Rightarrow \] \[{{10}^{2(bx+10)}}={{10}^{ax+cx+20}}\] \[\Rightarrow \] \[2(bx+10)=ax+cx+20,\ \,x\] \[\Rightarrow \] \[2b=a+c\ \ i.e.\ \ a,\ b,\ c\] 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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