Let starting term, a = 9
Also let a = \(t_{1}\)
So, \(t_{1}\) = 9
\(t_{2}\) = 15
\(t_{3}\) = 21
\(t_{2}\) - \(t_{1}\) = 6
\(t_{3}\) - \(t_{2}\) = 6
∴ this is an arithmetic sequence.
We need to formulate a rule of the form:
\(t_{n}\) = a + (n - 1)d
where n = the \(n^{th}\) term
So, let n = 32
\(t_{32}\) = 9 + (32 - 1)6
= 9 + 31 x 6
= 9 + 186
= 195
If I have calculated correctly, the 32nd term shoud be 195.
