If m times the mth term is equal to n times the nth term of an A.P. Prove that (m + w)th term of an A.P. is zero.

Solution

Let a be the first term and d be the common difference
Since m times the mth term is equal to n times the nth term.

∴                                $mt subscript straight m space equals space nt subscript straight n$
or                      m[a + (m - 1)d] = n[a + (n - 1)d]
or                      ma + m(m - 1)d = na + n (n-1)d
or ma - na + $left parenthesis straight m squared minus straight m right parenthesis straight d space minus space left parenthesis straight n squared minus straight n right parenthesis straight d space equals space 0$
or (m - n) a + ( $straight m squared minus straight m minus straight n squared plus straight n right parenthesis straight d space equals space 0$
or      (m - n) a + [(m² - n²) - (m - n)] d = 0
or (m - n) a + [(m - n) (m + n) - (m - n)]d = 0
or         (m - n) a + (m - n) (m + n - 1)d = 0
or (m - n) [a + (m + n - 1)d] = 0
or    a + (m + n - 1) d =0                                   (∵   $straight m not equal to straight n$ )
Hence, $straight t subscript straight m plus straight n end subscript space equals space 0$

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Permutations And Combinations

If m times the mth term is equal to n times the nth term of an A.P. Prove that (m + w)th term of an A.P. is zero.

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