10th Class Mathematics Solved Paper - Mathematics 2017 Outside Delhi Set-III

  • question_answer How many terms of an A.P. 9, 17, 25, .., must be taken to give a sum of 636?

    Answer:

    A.P. is \[9,17,25,.....,{{S}_{n}}=636\]
    \[a=9,d=17-9=8\]
                                        \[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\]
                                     \[636=\frac{n}{2}[18+8n-8]\]
                                        \[636=\frac{n}{2}[10+8n]\]
                                        \[636=n(5+4n)\]
                                        \[636=5n+4{{n}^{2}}\]
                        \[4{{n}^{2}}+5n-636=0\]
              \[4{{n}^{2}}+53n-48n-636=0\]
             \[n(4n+53)-12(4n+53)=0\]
                       \[(n-12)(4n+53)=0\]
                                     \[n-12=0\]\[\left( \because \,n\ne \frac{-53}{4}as\,\,n>0 \right)\]
                                            \[n=12\]


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