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JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    Let \[{{a}_{1}},\,\,{{a}_{2}},\,\,{{a}_{3}}\text{ }...\]be a G.P. such that\[{{a}_{1}}<0\], \[{{a}_{1}}+{{a}_{2}}=4\]and\[{{a}_{3}}+{{a}_{4}}=16\]. If \[\sum\limits_{i=1}^{9}{{{a}_{i}}}=4\lambda \], then \[\lambda \] is equal to [JEE MAIN Held on 07-01-2020 Evening]

    A) \[-\,513\]         

    B) \[-\,171\]

    C) \[\frac{511}{3}\]        

    D) \[171\]

    Correct Answer: B

    Solution :

    [b]
    Let the G.P. be \[a,\text{ }ar,\text{ }a{{r}^{2}},\,a{{r}^{3}},\,......\]and\[a<0\].
    \[\because \,\,\,\,\,\,\,\,{{a}_{1}}+{{a}_{2}}=4\Rightarrow a\,(1+r)=4\]               ...(i)
    \[{{a}_{3}}+{{a}_{4}}=16\Rightarrow a{{r}^{2}}\text{(}1+r\text{)}=16\]               ...(ii)
    \[\therefore \] from (i) and (ii), \[r=\pm \,2\].
    if r = 2, then \[a=\frac{4}{3}\]
    if r = -2, then a = -4.
    \[\therefore \,\,\sum\limits_{i=1}^{9}{{{a}_{i}}}=\frac{a({{r}^{9}}-1)}{r-1}=4\lambda \]
    \[=\frac{-\,4\cdot ({{(-2)}^{9}}-1)}{-\,2-1}=4\lambda \]
    \[\therefore \,\,\lambda \,=-\,171\]


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