[tex]n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\stackrel{\textit{term position}}{14}\\ a_1=\stackrel{\textit{first term}}{10}\\ d=\stackrel{\textit{common difference}}{-9} \end{cases} \\\\\\ a_{14}=10+(14-1)(-9)\implies a_{14}=10+(13)(-9)\implies a_{14}=-107[/tex]
Answer:
The 14th term of arithmetic sequence is -107.
Step-by-step explanation:
Here's the required formula to find the arithmetic sequence :
[tex]\longrightarrow{\pmb{\sf{a_n = a_1 + (n - 1)d}}}[/tex]
Substituting all the given values in the formula to find the 14th term of arithmetic sequence :
[tex]\begin{gathered} \qquad{\twoheadrightarrow{\sf{a_n = a_1 + (n - 1)d}}}\\\\\qquad{\twoheadrightarrow{\sf{a_{14} = 10 + (14 - 1) - 9}}}\\\\\qquad{\twoheadrightarrow{\sf{a_{14} = 10 + (13) - 9}}}\\\\\qquad{\twoheadrightarrow{\sf{a_{14} = 10 + 13 \times - 9}}}\\\\\qquad{\twoheadrightarrow{\sf{a_{14} = 10 - 117}}}\\\\\qquad{\twoheadrightarrow{\sf{a_{14} = - 107}}}\\\\\qquad{\star{\underline{\boxed{\sf{\pink{a_{14} = - 107}}}}}} \end{gathered}[/tex]
Hence, the 14th term of arithmetic sequence is -107.
[tex]\rule{300}{2.5}[/tex]
