Respuesta :
Given:
The sequence of series is given as 1/5, 23/15, 43/15, 21/5.....
The objective is to find the common difference and the next three terms of the series.
Explanation:
The difference can be calculated by subtracting the second term and the first term of the series in the sequence.
[tex]\begin{gathered} d=\frac{23}{15}-\frac{1}{5} \\ =\frac{23(5)-1(15)}{15(5)} \\ =\frac{115-15}{75} \\ =\frac{100}{75} \\ =\frac{4}{3} \end{gathered}[/tex]Now, the next three terms can be calculated by adding d to the fourth given term of the series.
[tex]\begin{gathered} a_5=\frac{21}{5}+\frac{4}{3} \\ =\frac{21(3)+4(5)}{5(3)} \\ =\frac{63+20}{15} \\ =\frac{83}{15} \end{gathered}[/tex]Now, the sixth term can be calculated as,
[tex]\begin{gathered} a_6=a_5+d \\ =\frac{83}{15}+\frac{4}{3} \\ =\frac{83(3)+4(15)}{15(3)} \\ =\frac{103}{15} \end{gathered}[/tex]The seventh term can be calculated as,
[tex]\begin{gathered} a_7=a_6+d \\ =\frac{103}{15}+\frac{4}{3} \\ =\frac{103(3)+4(15)}{15(3)} \\ =\frac{41}{5} \end{gathered}[/tex]Hence, the next three terms in the sequence are 83/15, 103/15, 41/5.
