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Exercise 8.1 · Q2

Q.Write the first five terms of the sequence whose nnth term is an=nn+1a_n = \dfrac{n}{n+1}.

Madhya Pradesh MpbseTextbookSubjective· 2mImportance★★★★★est
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✓ Free question

The sequence an=nn+1a_n = \frac{n}{n+1} is evaluated by substituting n=1,2,3,4,5n = 1, 2, 3, 4, 5 into the formula. The first five terms are 12,23,34,45,56\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}.

The idea here is straightforward: a sequence is just a list of numbers generated by a rule. The rule tells you what the nnth term is, so to get the first five terms, you plug in n=1n = 1, then n=2n = 2, and so on up to n=5n = 5. Each substitution gives you a fraction, and you simplify if needed.

Let’s walk through it step by step.

  1. For n=1n = 1:

    a1=11+1=12a_1 = \frac{1}{1+1} = \frac{1}{2}.

    This is the first term.

  2. For n=2n = 2:

    a2=22+1=23a_2 = \frac{2}{2+1} = \frac{2}{3}.

  3. For n=3n = 3:

    a3=33+1=34a_3 = \frac{3}{3+1} = \frac{3}{4}.

  4. For n=4n = 4:

    a4=44+1=45a_4 = \frac{4}{4+1} = \frac{4}{5}.

  5. For n=5n = 5:

    a5=55+1=56a_5 = \frac{5}{5+1} = \frac{5}{6}.

Tip

Notice a pattern: each term is of the form nn+1\frac{n}{n+1}, which is always less than 1 but gets closer to 1 as nn increases. For large nn, the terms approach 1 — that’s the limit of the sequence.

Watch out

A common mistake is to forget that the denominator is n+1n+1, not nn. For n=1n=1, some students write 11=1\frac{1}{1} = 1 instead of 12\frac{1}{2}. Always check the formula carefully.

So the first five terms are 12,23,34,45,56\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}.

✓Final answer

The first five terms are 12,23,34,45,56\boxed{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}}.

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