A) They are in AP
B) They are in GP
C) They are in HP
D) They are neither in AP, nor in GP, nor in HP
Correct Answer: B
Solution :
[b] The given equation \[({{a}^{2}}+{{b}^{2}}){{x}^{2}}-2b(a+c)x+({{b}^{2}}+{{c}^{2}})=0\] has equal roots, so, discriminant = 0 Hence, \[\{2b{{(a+c)}^{2}}-4({{a}^{2}}+{{b}^{2}})({{b}^{2}}+{{c}^{2}})=0\] \[\Rightarrow \,\,\,4{{b}^{2}}({{a}^{2}}+{{c}^{2}}+2ca)-4({{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{4}}+{{b}^{2}}{{c}^{2}})=0\]\[\Rightarrow \,\,{{b}^{2}}{{a}^{2}}+{{b}^{2}}{{c}^{2}}+2{{b}^{2}}ca-{{a}^{2}}{{b}^{2}}-{{a}^{2}}{{c}^{2}}\] \[-{{b}^{4}}-{{b}^{2}}{{c}^{2}}=0\] \[\Rightarrow \,\,\,\,2{{b}^{2}}ca={{b}^{4}}+{{a}^{2}}{{c}^{2}}\] \[\Rightarrow \,\,\,\,b4-2{{b}^{2}}ca+{{a}^{2}}{{c}^{2}}=0\] \[\Rightarrow \,\,\,\,{{({{b}^{2}}-ac)}^{2}}=0\] \[\Rightarrow \,\,\,\,{{b}^{2}}=ac\] \[\Rightarrow \] a, b, c are in GP.You need to login to perform this action.
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