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Exercise 5.1 · Q11

Q.If the sum of nn terms of an A.P. is 3n2+5n3n^2 + 5n, and its mmth term is 164, find the value of mm.

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Deriving the nnth term from SnS_n gives an=6n+2a_n=6n+2; solving 6m+2=1646m+2=164 gives m=27m=27.

For n≥2n\ge2: an=Sn−Sn−1a_n=S_n-S_{n-1} (and a1=S1a_1=S_1).

  1. Sn=3n2+5nS_n=3n^2+5n, so Sn−1=3(n−1)2+5(n−1)=3n2−6n+3+5n−5=3n2−n−2S_{n-1}=3(n-1)^2+5(n-1)=3n^2-6n+3+5n-5=3n^2-n-2.
  2. an=Sn−Sn−1=(3n2+5n)−(3n2−n−2)=6n+2a_n=S_n-S_{n-1}=(3n^2+5n)-(3n^2-n-2)=6n+2.
  3. Set am=164a_m=164: 6m+2=164⇒6m=162⇒m=276m+2=164\Rightarrow 6m=162\Rightarrow m=27. …

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