Question

Show that the sequence 9, 12, 15, 18, … is in A.P. Find its ${16^{th}}$term and the
 general term.

Answer 1

Hint: Use the common difference to prove it is an A.P. Find a pattern to find the
 general term and substitute for the ${16^{th}}$ term.

 The given sequence is 9, 12, 15, 18, …
To prove that the sequence is in A.P, the common difference between any two consecutive
 terms must be the same.
$\begin{gathered}
  12 - 9 = 3 \\
  15 - 12 = 3 \\
  18 - 15 = 3 \\
\end{gathered} $
Hence, the common difference is the same for all consecutive terms in the series. So, it is an
 A.P.
In an A.P., the first term is denoted as$a$.
$a = 9$
The common difference is denoted as$d$. It is the difference between any one term and its
 previous term. It is the same for when calculated for any term in the A.P. It is also used to find the next terms in the A.P.
$d = Tn + 1 - Tn = 12 - 9 = 3$
The ${n^{th}}$term $Tn = Tn - 1 + d$
Let us find the general term first.
$\begin{gathered}
  T1 = 9 \\
  T2 = 12 = 9 + 3\left( 1 \right) = 9 + 3\left( {2 - 1} \right) \\
  T3 = 15 = 9 + 3\left( 2 \right) = 9 + 3\left( {3 - 1} \right) \\
  T4 = 18 = 9 + 3\left( 3 \right) = 9 + 3\left( {4 - 1} \right) \\
\end{gathered} $
Hence, we can generalize this A.P. as $Tn = 9 + 3\left( {n - 1} \right)$ …(1)
We can substitute any value for $n$ to find the ${n^{th}}$ term.
The ${16^{th}}$term can be found by substituting $n = 16$in (1)
$\begin{gathered}
  Tn = 9 + 3\left( {n - 1} \right) \\
  T16 = 9 + 3\left( {16 - 1} \right) = 9 + 15\left( 3 \right) = 54 \\
\end{gathered} $
Note: The general term in an A.P. is found by finding the pattern in which the A.P. progresses.
 With the general term, any term in the A.P. can be found easily.