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JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    Given a sequence of 4 numbers, first  three of with are in G.P. and the last there are in A.P. with common difference six If first and last terms of this sequence are equal then the last term is :     JEE Main Online Paper ( Held On 25  April 2013 ) 

    A)                 16                                          

    B)                                        8

    C)                                        4                                             

    D)                                        2

    Correct Answer: B

    Solution :

                    Let a, b, c, d be four numbers of the sequence. Now, according to the question \[{{b}^{2}}=ac\] and c - b = 6 and a - c = 6 Also, given\[\] \[\therefore \]\[{{b}^{2}}=ac\Rightarrow {{b}^{2}}=a\left[ \frac{a+b}{2} \right]\]\[(\because 2c=a+b)\] \[\Rightarrow \]\[{{a}^{2}}-2{{b}^{2}}+ab=0\] Now,\[c-b=6\]and\[a-c=6,\]gives\[a-b=12\] \[\Rightarrow \]\[b=a-12\] \[\therefore \]\[{{a}^{2}}-2{{b}^{2}}+ab=0\] \[\Rightarrow \]\[{{a}^{2}}-2{{(a-12)}^{2}}+a(a-12)=0\] \[\Rightarrow \]\[{{a}^{2}}-2{{a}^{2}}-288+48a+{{a}^{2}}-12a=0\] \[\Rightarrow \]\[36a=288\]\[\Rightarrow \]\[a=8\] Hence, last term is d= a = 8.


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