Question 12

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

Answer

Let a and b be the first term and the common difference of the A.P. respectively.

According to the given condition,

$fraction numerator S u m space o f space m space t e r m s over denominator s u m space o f space n space t e r m s end fraction equals m squared over n squared rightwards double arrow fraction numerator begin display style m over 2 end style open square brackets 2 a plus open parentheses m minus 1 close parentheses d close square brackets over denominator begin display style n over 2 end style open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets end fraction equals m squared over n squared rightwards double arrow space space space fraction numerator 2 a plus left parenthesis m minus 1 right parenthesis d over denominator 2 a plus left parenthesis n minus 1 right parenthesis d end fraction equals m over n space space space space space.... left parenthesis 1 right parenthesis P u t t i n g space m space equals space 2 m space � space 1 space a n d space n space equals space 2 n space � space 1 space i n space left parenthesis 1 right parenthesis comma space w e space o b t a i n fraction numerator 2 a plus left parenthesis 2 m minus 2 right parenthesis d over denominator 2 a plus left parenthesis 2 n minus 2 right parenthesis d end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction rightwards double arrow fraction numerator a plus left parenthesis m minus 1 right parenthesis d over denominator a plus left parenthesis n minus 1 right parenthesis d end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction space space space space.... left parenthesis 2 right parenthesis fraction numerator m to the power of t h end exponent t e r m space o f space A. P. over denominator n to the power of t h space end exponent t e r m space o f space A. P. end fraction equals fraction numerator a plus left parenthesis m minus 1 right parenthesis d over denominator a plus left parenthesis n minus 1 right parenthesis d end fraction space space space space... left parenthesis 3 right parenthesis F r o m space left parenthesis 2 right parenthesis space a n d space left parenthesis 3 right parenthesis comma space w e space o b t a i n fraction numerator m to the power of t h end exponent t e r m space o f space A. P. over denominator n to the power of t h space end exponent t e r m space o f space A. P. end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction T h u s comma space t h e space g i v e n space r e s u l t space i s space p r o v e d.$