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Q.Find the first 5 terms of the sequence whose nnth term is an=2n−36a_n = \dfrac{2n - 3}{6}.

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2019Subjective· 2mImportance★★★★★
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Substituting n=1n=1 to 55 into an=2n−36a_n = \dfrac{2n-3}{6} gives −16,16,12,56,76-\dfrac{1}{6}, \dfrac{1}{6}, \dfrac{1}{2}, \dfrac{5}{6}, \dfrac{7}{6}.

We are given an=2n−36a_n = \dfrac{2n-3}{6}. Substituting each value of nn:

  • n=1n=1: a1=2(1)−36=−16a_1 = \dfrac{2(1)-3}{6} = \dfrac{-1}{6}
  • n=2n=2: a2=2(2)−36=16a_2 = \dfrac{2(2)-3}{6} = \dfrac{1}{6}
  • n=3n=3: a3=2(3)−36=36=12a_3 = \dfrac{2(3)-3}{6} = \dfrac{3}{6} = \dfrac{1}{2} …

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