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  • If m times mth term of an Arithmetic progression is equal to n times its nth term and m   n, show that the 

If m times mth term of an Arithmetic progression is equal to n times its nth term and m \neq  n, show that the \mathrm{(m+n)^{th}} is zero.

 

 
 
 
 
 

Answers (1)

    \mathrm{ma_m = na_n}

\mathrm{\Rightarrow ma + m(m-1)d = na + n(n-1)d}

\mathrm{\Rightarrow (m-n)a + (m^2 - m - n^2 + n)d = 0}

\mathrm{\Rightarrow (m-n)a + \left [(m-n)(m+n) - (m-n) \right ]d = 0}

Dividing by (m-n)

So,    \mathrm{ a + (m+n -1)d = 0}

                                      \mathrm{ a_{m+n} = 0}     

Posted by

Safeer PP

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