Solution:
Given:
[tex]108,105,102,...,15[/tex]
This is an arithmetic sequence because it has a common difference.
From the nth term of an arithmetic sequence,
[tex]\begin{gathered} a_n=a+(n-1)d \\ a=108 \\ d=105-108=102-105=-3 \\ a_n=15 \\ \\ Hence,\text{ the number of terms in the sequence is;} \\ 15=108+(n-1)-3 \\ 15-108=-3(n-1) \\ -93=-3(n-1) \\ \frac{-93}{-3}=n-1 \\ 31=n-1 \\ 31+1=n \\ 32=n \\ n=32 \end{gathered}[/tex]
The sum of an arithmetic sequence is given by;
[tex]\begin{gathered} S_n=\frac{n}{2}(a+l) \\ \\ where: \\ first\text{ term, }a=108 \\ last\text{ term, }l=15 \\ n=32 \\ \\ Hence, \\ S_n=\frac{32}{2}(108+15) \\ S_n=16(123) \\ S_n=16\times123 \\ S_n=1968 \end{gathered}[/tex]
herefore, the sum oft the sequence is 1968
