Question

Find the 14th

term of the arithmetic sequence whose common difference is d= -9 and whose first term is a, = 10.

Answer 1

`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`

Answer 2

The 14th term of arithmetic sequence is -107.

Explanation:

Here's the required formula to find the arithmetic sequence :

`\longrightarrow{\pmb{\sf{a_n = a_1 + (n - 1)d}}}`

• 
  `\blue\star` aₙ = nᵗʰ term in the sequence
• 
  `\blue\star` a₁ = first term in sequence
• 
  `\blue\star` n = number of terms
• 
  `\blue\star` d = common difference

Substituting all the given values in the formula to find the 14th term of arithmetic sequence :

• 
  `\green\star` aₙ = a₁₄
• 
  `\green\star` a₁ = 10
• 
  `\green\star` n = 14
• 
  `\green\star` d = -9

`\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 * - 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}`

Hence, the 14th term of arithmetic sequence is -107.

`\rule{300}{2.5}`