[tex]\bf 12~~,~~\stackrel{12+3}{15}~~,~~\stackrel{15+3}{18}~~,~~\stackrel{18+3}{21}~\hspace{10em}\stackrel{\textit{common difference}}{d=3} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} a_1=12\\ d=3 \end{cases} \\\\\\ a_n=12+(n-1)3\implies a_n=12+3n-3\implies a_n=3n+9[/tex]
Answer:
tₙ = 3(3 + n)
Step-by-step explanation:
Points to remember
nth term of an AP
tₙ = a + (n - 1)d
Where a - first term of AP
d - Common difference of AP
To find the nth term
The given series is,
12,15,18,21 .....
Here a = 12 and d = 15 - 12 = 3
tₙ = a + (n - 1)d
= 12 + (n - 1)3
=12 + 3n - 3
= 9 + 3n
= 3(3 + n)
Therefore tₙ = 3(3 + n)
