MATH SOLVE

4 months ago

Q:
# Please help I don’t really understand how to go about it.

Accepted Solution

A:

[tex]\bf \textit{sum of an arithmetic sequence}
\\\\
S_n=\cfrac{n(a_1+a_n)}{2}\qquad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
----------\\
S_n=795\\
a_1=102\\
a_n=57
\end{cases}
\\\\\\
795=\cfrac{n(102+57)}{2}\implies 1590=159n
\\\\\\
\cfrac{1590}{159}=n\implies 10=n\\\\
-------------------------------[/tex]

so the nth term is really the 10th term, and we know that's 57, thus

[tex]\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}\\ ----------\\ n=10\\ a_{10}=57\\ a_1=102 \end{cases} \\\\\\ 57=102+(10-1)d\implies 57=102+9d\implies -45=9d \\\\\\ \cfrac{-45}{9}=d\implies -5=d[/tex]

so, that's the common difference... .so you'd surely know what the 3rd term is, notice the first one is 102.

so the nth term is really the 10th term, and we know that's 57, thus

[tex]\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}\\ ----------\\ n=10\\ a_{10}=57\\ a_1=102 \end{cases} \\\\\\ 57=102+(10-1)d\implies 57=102+9d\implies -45=9d \\\\\\ \cfrac{-45}{9}=d\implies -5=d[/tex]

so, that's the common difference... .so you'd surely know what the 3rd term is, notice the first one is 102.