Find the 16" term of the arithmetic sequence whose common difference is d= 8 and whose first term is a, = 1.
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Answer:

[tex]a_{16} = 121[/tex].

Step-by-step explanation:

To go from the first term [tex]a_{1}[/tex] of this sequence to the second term [tex]a_{2}[/tex], add [tex]d[/tex] to [tex]a_{1}\![/tex].

To go from the first term [tex]a_{1}[/tex] of this sequence to the third term [tex]a_{3}[/tex], add [tex]2\, d[/tex] to [tex]a_{1}\![/tex].

In general, going from the first term [tex]a_{1}[/tex] of an arithmetic sequence to the [tex]k[/tex]th term, add [tex](k - 1)\, d[/tex] to [tex]a_{1}\![/tex].

Thus, given an arithmetic sequence with [tex]a_{1}[/tex] as the first term and [tex]d[/tex] as the common difference, the [tex]k[/tex]th term of this sequence would be [tex]a_{k} = a_{1} + (k - 1)\, d[/tex].

In this question, [tex]a_{1} = 1[/tex] and [tex]d = 8[/tex] for this arithmetic sequence. The [tex]16[/tex]th term of this sequence would be:

[tex]\begin{aligned}a_{16} &= 1 + (16 - 1) \times 8 \\ &= 1 + 15 \times 8 \\ &= 121\end{aligned}[/tex].

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