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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    Let\[{{S}_{n}}\]denote the sum of first n terms of an AP and\[{{S}_{2n}}=3{{S}_{n}}.\]If\[{{S}_{3n}}=k{{S}_{n}},\]then the value of k is equal to

    A)  4                                            

    B)  5

    C)  6                            

    D)         7

    E)  8

    Correct Answer: C

    Solution :

    Given, \[{{S}_{2n}}=3{{S}_{n}}\] \[\frac{2n}{2}[2a+(2n-1)d]=.3\frac{n}{2}[2a+(n-1)d]\] Where, a and d are first term and common difference of an AP respectively. \[\Rightarrow \]               \[4a+2(2n-1)d=6a+3(n-1)d\] \[\Rightarrow \]               \[2a+(3n-3-4n+2)d=0\] \[\Rightarrow \]               \[2a+(-n-1)d=0\] \[\Rightarrow \]               \[2a+(n+1)(-d)=0\] \[\Rightarrow \]               \[2a=(n+1)d\]                    ?..(i) Now, \[\frac{{{S}_{3n}}}{{{S}_{n}}}=\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}\]                 \[=\frac{3[(n+1)d+(3n-1)d]}{[(n+1)d+(n-1)d]}\][from Eq. (i)]                 \[=\frac{3[(n+1+3n-1)d]}{(n+1+n-1)d}=\frac{3(4nd)}{(2nd)}=6\] \[\Rightarrow \]                               \[{{S}_{3n}}=6{{S}_{n}}\] On compare with, \[{{S}_{3n}}=k{{S}_{n}}\] \[\Rightarrow \]                               \[k=6\]


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