(A) Given the $n^{th}$ term $a_{n} = \frac{n}{n+1}$.Substituting $n = 1, 2, 3, 4, 5$,we obtain:$a_{1} = \frac{1}{1+1} = \frac{1}{2}$$a_{2} = \frac{2}{

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$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$

(A) Given the $n^{th}$ term $a_{n} = \frac{n}{n+1}$.Substituting $n = 1, 2, 3, 4, 5$,we obtain:$a_{1} = \frac{1}{1+1} = \frac{1}{2}$$a_{2} = \frac{2}{2+1} = \frac{2}{3}$$a_{3} = \frac{3}{3+1} = \frac{3}{4}$$a_{4} = \frac{4}{4+1} = \frac{4}{5}$$a_{5} = \frac{5}{5+1} = \frac{5}{6}$Therefore,the first five terms are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

Easy

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = \frac{n}{n+1}$.

A
$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$
B
$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{6}{7}$
C
$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{5}{6}, \frac{6}{7}$
D
$\frac{1}{2}, \frac{2}{3}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}$

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nth term of special series, Sum to n terms and Infinite number of terms

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Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = \frac{n}{n+1}$.

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