10th Class Mathematics Solved Paper - Mathematics-2016 Outside Delhi Set-I

  • question_answer The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

    Answer:

    We know that
                                        \[{{T}_{n}}=a+(n-1)d\]
    Given,                           \[{{T}_{4}}=a+(4-1)d=0\]
                                   \[a+3d=0\]
                                          \[a=-3d\]
                                        \[{{T}_{25}}=a+(25-1)d\]
                                             \[=a+24d=(-3d)+24d\]
                                             \[=24d\]
    And,                             \[{{T}_{11}}=a+(11-1)d\]
                                            \[=a+10d\]
    Then,                            \[3{{T}_{11}}=3(a+10d)\]
                                            \[=3a+30d\]
                                            \[=3(-3d)+30d(a=-3d)\]
                                            \[=30d-9d=21d={{T}_{25}}\]
    \[\therefore \]                              \[3{{T}_{11}}={{T}_{25}}\]                Hence Proved.


adversite


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