Write first five terms of sequences whose n^{t h} term is a_{n}=frac{2 n-3}{6}

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8. Sequences and Series

easy

Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=\frac{2 n-3}{6}$

A

$\frac{{ - 1}}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$

B

$\frac{{ - 1}}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$

C

$\frac{{ - 1}}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$

D

$\frac{{ - 1}}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$

Solution

Substituting $n=1,2,3,4,5,$ we obtain

$a_{1}=\frac{2 \times 1-3}{6}=\frac{-1}{6}$

$a_{2}=\frac{2 \times 2-3}{6}=\frac{1}{6}$

$a_{3}=\frac{2 \times 3-3}{6}=\frac{3}{6}=\frac{1}{2}$

$a_{4}=\frac{2 \times 4-3}{6}=\frac{5}{6}$

$a_{5}=\frac{2 \times 5-3}{6}=\frac{7}{6}$

Therefore, the required terms are $\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}$ and $\frac{7}{6}$

Standard 11

Mathematics

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