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Exercise 5.1 · Q19

Q.If the ratio of the sum of the first mm terms of an A.P. to the sum of the first nn terms is m2:n2m^2:n^2, find the ratio of its 19th term to its 21st term.

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The given ratio of sums forces d=2ad=2a in the A.P., which yields a19:a21=37:41a_{19}:a_{21}=37:41.

Sum of nn terms: Sn=n2[2a+(n−1)d]S_n=\dfrac n2\big[2a+(n-1)d\big]. nnth term: an=a+(n−1)da_n=a+(n-1)d.

  1. SmSn=m2[2a+(m−1)d]n2[2a+(n−1)d]=m[2a+(m−1)d]n[2a+(n−1)d]=m2n2\dfrac{S_m}{S_n}=\dfrac{\frac m2[2a+(m-1)d]}{\frac n2[2a+(n-1)d]}=\dfrac{m[2a+(m-1)d]}{n[2a+(n-1)d]}=\dfrac{m^2}{n^2} (given).
  2. This simplifies to 2a+(m−1)d2a+(n−1)d=mn\dfrac{2a+(m-1)d}{2a+(n-1)d}=\dfrac mn for every choice of m,nm,n.
  3. Define f(k)=2a+(k−1)df(k)=2a+(k-1)d. The relation f(m)f(n)=mn\dfrac{f(m)}{f(n)}=\dfrac mn for all m,nm,n means f(k)=ckf(k)=ck for some constant cc.
  4. At k=1k=1: f(1)=2a=c(1)⇒c=2af(1)=2a=c(1)\Rightarrow c=2a. So 2a+(k−1)d=2ak2a+(k-1)d=2ak for every kk.
  5. Taking k=2k=2: 2a+d=4a⇒d=2a2a+d=4a\Rightarrow d=2a. …

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