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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    Let \[{{a}_{n}}\] be the \[{{n}^{th}}\] term of an A.P. If \[\sum\limits_{r=1}^{100}{{{a}_{2r}}=\alpha }\] and \[\sum\limits_{r=1}^{100}{{{a}_{2r}}-1=\beta },\] then the common difference of the A.P. is     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) \[\alpha -\beta \]

    B) \[\frac{\alpha -\beta }{100}\]

    C) \[\beta -\alpha \]

    D) \[\frac{\alpha -\beta }{200}\]

    Correct Answer: B

    Solution :

                    Let A.P. be a, a + d, a + 2d, ....... \[{{a}_{2}}+{{a}_{4}}+.....+{{a}_{200}}=\alpha \] \[\Rightarrow \]\[\frac{100}{2}[2(a+d)+(100-1)d]=\alpha \]                                                         ?(i) and\[{{a}_{1}}+{{a}_{3}}+{{a}_{5}}+.....+{{a}_{199}}=\beta \] \[\Rightarrow \]\[\frac{100}{2}[2a+(100-1)d]=\beta \]                                                                    ?.(ii) on solving (i) and (ii) \[d=\frac{\alpha -\beta }{100}\]


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