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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    A student read common difference of an AP as\[-3\]instead of 3 and obtained the sum of first 10 terms as\[-30\]. Then, the actual sum of first 10 terms is equal to

    A)  240       

    B)                         120

    C)  300                       

    D)         180

    E)  480

    Correct Answer: A

    Solution :

    Let common difference\[{{d}_{1}}=-3\]and first term be a. \[\therefore \]Series become \[a,\text{ }a-3,\text{ }a-6,...,\text{ }a-27\] \[\therefore \]  \[S=10a+(-3-6-....-27)\] \[\Rightarrow \]               \[-30=10a-3(1+2+...+9)\] \[\Rightarrow \]               \[-30=10a-3\left[ \frac{9(9+1)}{2} \right]\] \[\Rightarrow \]               \[-30=10a-135\] \[\Rightarrow \]               \[10a=105\] \[\Rightarrow \]               \[a=\frac{105}{10}\] Now, correct common difference\[{{d}_{2}}=3\] \[\therefore \]Required sum \[=\frac{10}{2}\left[ 2\times \frac{105}{10}+(10-1)3 \right]\]                                 \[=5\left[ \frac{105}{5}+27 \right]\]                                 \[=5\times 48=240\]


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