A) 1 : 2 : 3
B) 1 : 3 : 4
C) 2 : 3 : 4
D) 1 : 2 : 4
Correct Answer: A
Solution :
Given that, a, b, c are in AP \[\Rightarrow \] \[2b=a+c\] ?(i) and \[b-a,c-b,a\]are in GP \[\Rightarrow \] \[{{(c-b)}^{2}}=(b-a)a\] \[\Rightarrow \] \[{{c}^{2}}+{{b}^{2}}-2bc=ab-{{a}^{2}}\] \[\Rightarrow \] \[{{c}^{2}}+{{\left( \frac{a+c}{2} \right)}^{2}}-2\left( \frac{a+c}{2} \right)c\] \[=a\left( \frac{a+c}{2} \right)-{{a}^{2}}\][Using Eq.(i)] \[\Rightarrow \]\[{{c}^{2}}+\frac{{{a}^{2}}+{{c}^{2}}+2ac}{4}-ac-{{c}^{2}}=\frac{{{a}^{2}}}{2}+\frac{ac}{2}-{{a}^{2}}\] \[\Rightarrow \] \[\frac{{{a}^{2}}+{{c}^{2}}+2ac-4ac}{4}=\frac{ac-{{a}^{2}}}{2}\] \[\Rightarrow \] \[{{(c-a)}^{2}}=2a(c-a)\] \[\Rightarrow \] \[c-a=2a\] \[\Rightarrow \] \[c=3a\] From Eq. (i), \[2b=a+3a\] \[b=2a\] \[\therefore \] \[a:b:c=a:2a:3a\] \[=1:2:3\]
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