Permutations and Combinations

Permutations and Combinations

Question

The ratio of sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of with and nth term is 2m – 1 :2n – 1.

Answer

Let a be the first term and d be the common ratio.
space space space space space space fraction numerator Sum space of space first space straight m space terms space of space an space straight A. straight P. over denominator Sum space of space first space straight n space terms space of space an space straight A. straight P. end fraction space equals space straight m squared over straight n squared    or       fraction numerator begin display style straight m over 2 end style left square bracket 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d right square bracket over denominator begin display style straight n over 2 end style left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d end fraction equals space straight m squared over straight n squared
or            fraction numerator left square bracket 2 straight a space plus space left parenthesis straight m minus 1 right parenthesis straight d right square bracket over denominator left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket end fraction space equals space straight m over straight n              or      2an + n (m - 1)d = 2 am + m(n - 1)d
or           2am - 2an = (mn - n) d - (mn - m)d
or           2a (m - n) = d [mn -n -mn + m]
or           2a(m - n) = d (m -n)      or          d = 2a
           fraction numerator mth space term space of space an space straight A. straight P. over denominator nth space term space of space an space straight A. straight P. end fraction space equals space fraction numerator straight a plus left parenthesis straight m minus 1 right parenthesis straight d over denominator straight a plus left parenthesis straight n minus 1 right parenthesis straight d end fraction space equals space fraction numerator straight a plus left parenthesis straight m minus 1 right parenthesis 2 straight a over denominator straight a plus left parenthesis straight n minus 1 right parenthesis 2 straight a end fraction space equals space fraction numerator straight a left square bracket 1 plus 2 straight m minus 2 right square bracket over denominator straight a left square bracket 1 plus 2 straight n minus 2 right square bracket end fraction space equals space fraction numerator 2 straight m minus 1 over denominator 2 straight n minus 1 end fraction
∴     The ratio of mth term and nth term is:
                                  2m - 1 : 2n - 1

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