Respuesta :
Given:
The sequence is
[tex]a_n=\frac{3}{n+1}[/tex]Required:
Find the first term, the second term, the third term, the fourth term and the 100th term.
Explanation:
The given sequence is:
[tex]a_{n}=\frac{3}{n+1}[/tex]Substitute n = 1
[tex]\begin{gathered} a_1=\frac{3}{1+1} \\ a_1=\frac{3}{2} \end{gathered}[/tex]Substitute n = 2
[tex]\begin{gathered} a_2=\frac{3}{2+1} \\ a_2=\frac{3}{3} \\ a_2=1 \end{gathered}[/tex]Substitute n = 3
[tex]\begin{gathered} a_3=\frac{3}{3+1} \\ a_3=\frac{3}{4} \end{gathered}[/tex]Substitute n = 4
[tex]\begin{gathered} a_4=\frac{3}{4+1} \\ a_4=\frac{3}{5} \end{gathered}[/tex]The term of the sequence are:
[tex]\frac{3}{2},1,\frac{3}{4},\frac{3}{5}[/tex]The given series is in HP
We will write it in AP as:
[tex]\frac{2}{3},1,\frac{4}{3},\frac{5}{3}[/tex]So the common difference of the given sequence is:
[tex]\begin{gathered} 1-\frac{2}{3}=\frac{1}{3} \\ \frac{4}{3}-1=\frac{1}{3} \\ \frac{5}{3}-\frac{4}{3}=\frac{1}{3} \end{gathered}[/tex]The nth term of the AP series is given by the formula:
[tex]a_n=a+(n-1)d[/tex]where a = first term
n = number of terms
d = common difference
[tex]\begin{gathered} a_{100}=\frac{2}{3}+(100-1)\times\frac{1}{3} \\ a_{100}=\frac{2}{3}+99\times\frac{1}{3} \\ a_{100}=\frac{2+99}{3} \\ a_{100}=\frac{101}{3} \end{gathered}[/tex]This is the 100th term for the AP.
The 100th term of the given HP sequence is:
[tex]\frac{3}{101}[/tex]Final Answer:
[tex]\begin{gathered} First\text{ term = }\frac{3}{2} \\ Second\text{ term = 1} \\ Third\text{ term = }\frac{3}{4} \\ Fourth\text{ term = }\frac{3}{5} \\ 100th\text{ term = }\frac{3}{101} \end{gathered}[/tex]