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JEE Main & Advanced Mathematics Sequence & Series Question Bank Arithmetic Progression

  • question_answer
    If \[a,\ b,\ c\] are in A.P., then \[\frac{{{(a-c)}^{2}}}{({{b}^{2}}-ac)}=\] [Roorkee 1975]

    A) 1

    B) 2

    C) 3

    D) 4

    Correct Answer: D

    Solution :

    If \[a,\ b,\ c\] are in A.P. \[\Rightarrow \]\[2b=a+c\] So, \[\frac{{{(a-c)}^{2}}}{({{b}^{2}}-ac)}=\frac{{{(a-c)}^{2}}}{\left\{ {{\left( \frac{a+c}{2} \right)}^{2}}-ac \right\}}\]                     \[=\frac{{{(a-c)}^{2}}4}{[{{a}^{2}}+{{c}^{2}}+2ac-4ac]}=\frac{4{{(a-c)}^{2}}}{{{(a-c)}^{2}}}=4\]. Trick: Put\[a=1,\ b=2,\ c=3\], then the required value is\[\frac{4}{1}=4\].


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