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11th Class Mathematics Sequence and Series Question Bank Sequence and Series (A.P., G.P. and H.P.)

  • question_answer
    For a sequence in AP, the sum of n terms of the sequence is \[3n-{{n}^{2}}\], then find the 101st term of the sequence,                                                               

    A)  201                             

    B)  - 201              

    C)  - 198   

    D)  + 198         

    Correct Answer: C

    Solution :

    (c): We have, \[Sn=3n-{{n}^{2}}\]        \[\] \[{{S}_{1}}=2,\text{ }{{S}_{2}}=2;\text{ }{{S}_{3}}=0;\text{ }{{S}_{4}}=-4;\text{ }{{S}_{5}}=-10\] Thus, common difference is d = ? 2 Hence the 101st term of the sequence is \[=a+\left( 101-1 \right)d=2+100\times \left( -2 \right)\text{ }=-198\]                     


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