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

  • question_answer If the \[{{p}^{th}}\] term of an A. P. is q and \[{{q}^{th}}\] term is p, prove that its \[{{n}^{th}}\] term is \[(p+q-n)\].

    Answer:

    Let a be first term and d be common difference.
    Then,                \[{{p}^{th}}\text{term}=q\Rightarrow a+(p-1)d=q\]                               ?(i)       
                            \[{{q}^{th}}\text{term}=p\Rightarrow a+(q-1)d=p\]                               ?(ii)
    On subtracting eq. (ii) from eq. (i)
               \[(p-1)d-(q-1)d=q-p\]
                  \[pd-d-qd+d=q-p\]
         \[(p-q)d=q-p\] or \[d=\frac{q-p}{p-q}=-1\]
    Putting value of d in eq. (i)
                \[a+(p-1)(-1)=q\]
                                 \[a=q+p-1\]
                         \[{{n}^{th}}term=a+(n-1)d\]
                            \[=q+p-1+(n-1)(-1)\]
                            \[=q+p-1+1-n=q+p-n\]
                         \[{{T}_{n}}=q+p-n\]                               Hence Proved
     


adversite


You need to login to perform this action.
You will be redirected in 3 sec spinner

Free
Videos