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Exercise 8.1 · Q4

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

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The sequence is defined by a linear expression in nn, so we evaluate it at n=1,2,3,4,5n = 1, 2, 3, 4, 5 to get the first five terms: −16,16,12,56,76-\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}.

The idea here is straightforward: a sequence is just a list of numbers generated by plugging in consecutive natural numbers (n=1,2,3,…n = 1, 2, 3, \dots) into the given formula for the nnth term. The formula an=2n−36a_n = \frac{2n-3}{6} is linear in nn, so the terms will increase steadily. There’s no trick — just careful arithmetic with fractions.

Let’s work through it step by step.

  1. For n=1n = 1: Substitute n=1n = 1 into ana_n:

a1=2(1)−36=2−36=−16a_1 = \frac{2(1) - 3}{6} = \frac{2 - 3}{6} = \frac{-1}{6}

So the first term is −16-\frac{1}{6}.

  1. For n=2n = 2:

a2=2(2)−36=4−36=16a_2 = \frac{2(2) - 3}{6} = \frac{4 - 3}{6} = \frac{1}{6}

The second term is 16\frac{1}{6}.

  1. For n=3n = 3:

a3=2(3)−36=6−36=36=12a_3 = \frac{2(3) - 3}{6} = \frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2}

Notice we simplify 36\frac{3}{6} to 12\frac{1}{2} — always reduce fractions when possible.

  1. For n=4n = 4:

a4=2(4)−36=8−36=56a_4 = \frac{2(4) - 3}{6} = \frac{8 - 3}{6} = \frac{5}{6}

This fraction is already in simplest form.

  1. For n=5n = 5:

a5=2(5)−36=10−36=76a_5 = \frac{2(5) - 3}{6} = \frac{10 - 3}{6} = \frac{7}{6}

This is an improper fraction, but that’s perfectly fine — we leave it as is. …

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