Question

What is a30 for the arithmetic sequence presented in the table below?

n
6
11
an
50
35


Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference.

Answer 1

`\bf \begin{array}{ll} n&a_n\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 6&50\\ 7&50+d\\ 8&50+d+d\\ 9&50+d+d+d\\ 10&50+d+d+d+d\\ 11&50+d+d+d+d+d\\ &35 \end{array} \\\\\\ 50+5d=35\implies 5d=-15\implies d=\cfrac{-15}{5}\implies d=-3 \\\\\\ \textit{we know d = -3, and we know }a_(11)=35\qquad thus`


`\bf n^(th)\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ d=-3\\ n=11\\ a_(11)=35 \end{cases} \\\\\\ a_(11)=a_1+(11-1)(-3)\implies 35=a_1+(11-1)(-3) \\\\\\ 35=a_1-30\implies 65=a_1`

now we know what "d" is, and what a₁ is, so let's check the 30th term,


`\bf n^(th)\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ d=-3\\ n=30\\ a_(1)=65 \end{cases} \\\\\\ a_(30)=65+(30-1)(-3)\implies a_(30)=65-87\implies a_(30)=-22`