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Consider the sequence 3, 6, 9, 12, 15, ..
(1) Write down the next two terms of the sequence.
(ii) Find, in terms of n, a formula for the nth term of the sequence.
(iii) Hence, find the 105th term.
$3,6,9,12,15, \ldots \ldots \ldots \ldots \ldots$
Let $\left\{a_{n}\right\}$ be the given sequence
First Term $=3$
Common Difference $=6-3=3$
So the General Term
$=n^{t h}$ term
$=a_{n}$
$=3+(n-1) \times 3$
$=3 n \quad$ where $n \in \mathbb{N}$
1. Next two Terms are
$a_{6}=3 \times 6=18$
$a_{7}=3 \times 7=21$
2. The formula for the nth term of the sequence $=a_{n}=3 n$
3. 105th term is $=a_{105}=3\times105$
= 315
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