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Find the first five terms of the sequence in which the $n^{th}$ term is given by (i) $t_n = 3n - 4$, (ii) $t_n = 2 - 5n$, and (iii) $t_n = n^2 - 2n + 3$ for $n \ge 1$.
Answer:
Find the $10^{th}$ and $15^{th}$ terms of the sequence $t_n = 5n - 3$ for $n \ge 1$.
Answer:
For the $10^{th}$ term, substitute $n = 10$: $t_{10} = 5(10) - 3 = 50 - 3 = \mathbf{47}$.
For the $15^{th}$ term, substitute $n = 15$: $t_{15} = 5(15) - 3 = 75 - 3 = \mathbf{72}$.
Determine whether 97 and 172 are terms of the sequence $t_n = 5n - 3$ for $n \ge 1$.
Answer:
Set $5n - 3 = 97 \implies 5n = 100 \implies n = 20$. Since 20 is a natural number, 97 is a term (the $20^{th}$ term).
Set $5n - 3 = 172 \implies 5n = 175 \implies n = 35$. Since 35 is a natural number, 172 is a term (the $35^{th}$ term).
Which term of the sequence $t_n = 5n - 3$ for $n \ge 1$ is 607?
Answer:
Set $5n - 3 = 607 \implies 5n = 610 \implies n = 122$.
607 is the $122^{nd}$ term of the sequence.
A sequence is given by the recursive rule $t_1 = -5, t_{n+1} = t_n + 3$ for $n \ge 1$. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
Answer:
The first five terms are: -5, -2, 1, 4, 7.
The explicit rule for this AP is $t_n = -5 + (n-1)3 = 3n - 8$.
Set $3n - 8 = 52 \implies 3n = 60 \implies n = 20$.
Yes, 52 is a term of this sequence; it is the $20^{th}$ term.
Let $T_1 = 1, T_2 = 2, T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \ge 4$. Find $T_4, T_5, T_6, T_7$, and $T_8$.
Answer:
$T_4 = 1 + 2 + 4 = \mathbf{7}$
$T_5 = 2 + 4 + 7 = \mathbf{13}$
$T_6 = 4 + 7 + 13 = \mathbf{24}$
$T_7 = 7 + 13 + 24 = \mathbf{44}$
$T_8 = 13 + 24 + 44 = \mathbf{81}$
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