The ratio of the sums of m and | Class 11 Mathematics Chapter Sequence and Series, Sequence and Series NCERT Solutions

Question:

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.


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Comments

  • Ajsgs
  • 2019-07-20 19:18:02

It is a concept of gp


  • Subham Kumar
  • 2019-02-24 10:51:05

Why we put m=2m-1 and n=2n-1 in (1)


  • Rehina
  • 2018-11-28 19:40:02

Thanks for your tip thnq soooo much


  • Lucky
  • 2017-01-12 05:54:55

Good question


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 11 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 12: The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that....