Question

Find the common difference and the next three terms in the sequence after the last one given. 1/5, 23/15, 43/15, 21/5 , ….

Answer 1

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.

`\begin{gathered} d=(23)/(15)-(1)/(5) \\ =(23(5)-1(15))/(15(5)) \\ =(115-15)/(75) \\ =(100)/(75) \\ =(4)/(3) \end{gathered}`

Now, the next three terms can be calculated by adding d to the fourth given term of the series.

`\begin{gathered} a_5=(21)/(5)+(4)/(3) \\ =(21(3)+4(5))/(5(3)) \\ =(63+20)/(15) \\ =(83)/(15) \end{gathered}`

Now, the sixth term can be calculated as,

`\begin{gathered} a_6=a_5+d \\ =(83)/(15)+(4)/(3) \\ =(83(3)+4(15))/(15(3)) \\ =(103)/(15) \end{gathered}`

The seventh term can be calculated as,

`\begin{gathered} a_7=a_6+d \\ =(103)/(15)+(4)/(3) \\ =(103(3)+4(15))/(15(3)) \\ =(41)/(5) \end{gathered}`

Hence, the next three terms in the sequence are 83/15, 103/15, 41/5.