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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Morning)

  • question_answer
    The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is        [JEE Main Online Paper Held On 12-Jan-2019 Morning]       

    A) 36                                

    B) 24   

    C) 28                                

    D)   32   

    Correct Answer: C

    Solution :

    Let 3 consecutive terms are \[\frac{a}{r},\,\,a,\,\,ar\]of a G.P. \[\therefore \]\[{{a}^{3}}=512\Rightarrow a=8\] Now,\[\frac{8}{r}+4,12,8r\]are in A.P. \[\therefore \]\[24=\frac{8}{r}+4+8r\Rightarrow 24r=8+4r+8{{r}^{2}}\] \[\Rightarrow \]\[2{{r}^{2}}-5r+2=0\Rightarrow (r-2)(2r-1)=0\] \[\Rightarrow \]\[r=2\]or\[\frac{1}{2}\] When r = 2, terms are 4, 8, 16 When \[r=\frac{1}{2},\]terms are 16, 8, 4 So, required sum \[=16+8+4=28\]


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