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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    If\[{{S}_{n}}\]denotes the sum of n terms of an AP, then\[{{S}_{n+3}}-3{{S}_{n+2}}+3{{S}_{n+1}}-{{S}_{n}}\]is equal to:

    A)  3      

    B)                                         1                            

    C)         \[\frac{1}{2}\]                                  

    D)                         2

    E)         0

    Correct Answer: D

    Solution :

    \[({{S}_{n+3}}-{{S}_{n+2}})-2({{S}_{n+2}}-{{S}_{n+1}})({{S}_{n+1}}-{{S}_{n}})\] \[{{t}_{n+3}}-2{{t}_{n+2}}+{{t}_{n+1}}\] Let a and d be first term and common difference of an AP respectively \[\therefore \]\[{{t}_{n+3}}-2{{t}_{n+2}}{{t}_{n+1}}=a+(n+3-1)d\]                 \[-2\{a+(n+2-1)d\}+a+(n+1-1)d\] \[=2a-2a+d(n+2-2n-2+n)\] \[=0+d(0)=0\] \[\therefore \]\[{{S}_{n+3}}-3{{S}_{n+2}}+3{{S}_{n+1}}-{{S}_{n}}=0\]


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