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Find the 14th
term of the arithmetic sequence whose common difference is d= -9 and whose first term is a, = 10.

Sagot :

[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]

AadhPritidn

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]

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

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

  • [tex]\green\star[/tex] aₙ = a₁₄
  • [tex]\green\star[/tex] a₁ = 10
  • [tex]\green\star[/tex] n = 14
  • [tex]\green\star[/tex] d = -9

[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]

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