Question
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$\text{a}_\text{n}=\frac{\text{n}-3}{3}$
$\text{a}_\text{n}=\frac{\text{n}-3}{3}$
✓
Answer
$\text{a}_\text{n}=\frac{\text{n}-3}{3}$
Here, the $n^{th}$ term is given by the above expression. So, to find the first term we use, $n = 1$, we get,
$\text{a}_1=\frac{(1)-2}{3}$
$=\frac{-1}{3}$
Similarly, we find the other four terms,
Second term (n = 2),
$\text{a}_1=\frac{(2)-2}{3}$
$=\frac{0}{3}$
$=0$
Third term (n = 3),
$\text{a}_3=\frac{(3)-2}{3}$
$=\frac{1}{3}$
Fourh terms (n = 4),
$\text{a}_4=\frac{(4)-2}{3}$
$=\frac{2}{3}$
Fifth term (n = 5),
$\text{a}_5=\frac{(5)-2}{3}$
$=\frac{3}{3}$
$=1$
Therefore, the first five terms for the given sequence are $\text{a}_1=\frac{-1}{3},\ \text{a}_2=0,\ \text{a}_3=\frac{1}{3},\ \text{a}_4=\frac{2}{3},\ \text{a}_5=1.$
Here, the $n^{th}$ term is given by the above expression. So, to find the first term we use, $n = 1$, we get,
$\text{a}_1=\frac{(1)-2}{3}$
$=\frac{-1}{3}$
Similarly, we find the other four terms,
Second term (n = 2),
$\text{a}_1=\frac{(2)-2}{3}$
$=\frac{0}{3}$
$=0$
Third term (n = 3),
$\text{a}_3=\frac{(3)-2}{3}$
$=\frac{1}{3}$
Fourh terms (n = 4),
$\text{a}_4=\frac{(4)-2}{3}$
$=\frac{2}{3}$
Fifth term (n = 5),
$\text{a}_5=\frac{(5)-2}{3}$
$=\frac{3}{3}$
$=1$
Therefore, the first five terms for the given sequence are $\text{a}_1=\frac{-1}{3},\ \text{a}_2=0,\ \text{a}_3=\frac{1}{3},\ \text{a}_4=\frac{2}{3},\ \text{a}_5=1.$
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