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Sagot :
The series is an arithmetic series so it is given that:
[tex]\begin{gathered} a_1=23,a_2=114 \\ d=a_2-a_1 \\ d=114-23=91 \end{gathered}[/tex]The nth term of the series is given by:
[tex]\begin{gathered} a_n=a_1+(n-1)(d) \\ a_n=23+(n-1)91 \\ a_n=23+91n-91 \\ a_n=91n-68\ldots(i) \end{gathered}[/tex]The (n-1)th term is given by:
[tex]\begin{gathered} a_{n-1}=91(n-1)-68 \\ a_{n-1}=91n-91-68 \\ a_{n-1}=91n-159\ldots(ii) \end{gathered}[/tex]Subtract (ii) from (i) to get:
[tex]\begin{gathered} a_n-a_{n-1}=-68-(-159) \\ a_n-a_{n-1}=91 \\ a_n=a_{n-1}+91\ldots(iii) \end{gathered}[/tex]So the recursive formula is given by equation (iii) shown above.
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