An arithmetic sequence is such that each term in the sequence is found by adding the same number to

Question

An arithmetic sequence is such that each term in the sequence is found by adding the same number to the previous term. The first five terms of an arithmetic sequence are given by $a$, $1$, $b$, $c$, $13$, …. Find the values of $a$, $b$ and $c$. Find an expression, in terms of $n$, for the ${{n}^{\mathrm{th}}}$ term, ${{T}_{n}}$, of this arithmetic sequence. Explain why the terms of the sequence are all odd numbers. Observe the following difference of unit fractions. ${\textbf{1}^{\mathrm{st}}\ \mathrm{line:}\ 1 - \frac{1}{2} = \frac{1}{2}}$ ${\textbf{2}^{\mathrm{nd}}\ \mathrm{line:}\ \frac{1}{2} - \frac{1}{3} = \frac{1}{6}}$ ${\textbf{3}^{\mathrm{rd}}\ \mathrm{line:}\ \frac{1}{3} - \frac{1}{4} = \frac{1}{12}}$ ${\vdots}$ ${\textbf{10}^{\mathrm{th}}\ \mathrm{line:}\ \frac{1}{10} - \frac{1}{11} = \frac{1}{110}}$ ${\vdots}$ Write down the ${{4}^{\mathrm{th}}}$ line and the ${{5}^{\mathrm{th}}}$ line. Write down an expression of the ${{n}^{\mathrm{th}}}$ line. Does the term $\frac{1}{421}$ occur in row $n$? Justify your answers with working. By adding the first $3$ lines, we obtain $1-\frac{1}{4}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}$. Thus, $1=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}$. Using this information, express 1 as a sum of seven different unit fractions.

Accepted Answer

(a)(i) $a=-3,b=5,c=9$. (ii) $T_n=4n-7$. (iii) Every term is odd. (b)(i) $1/4-1/5=1/20$, $1/5-1/6=1/30$. (ii) $1/n-1/(n+1)=1/[n(n+1)]$. (iii) No. (iv) $1=1/2+1/6+1/7+1/12+1/20+1/30+1/42$.

IP1U7 Exercise Q9

IP1U7 Exercise Q9

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