Exercise 8.1 Practice
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Overview
This page provides comprehensive Ch 8: Exploring Sequences and Progressions - Exercise Set 8.1 Practice. Practice generating sequence terms from explicit and recursive rules, testing value membership, and solving multi-step recursive relations with step-by-step solutions.
Explicit & Recursive Sequence Rules, Term Analysis & Tribonacci Sequences
Q1: First Five Terms of Sequences
Find the first five terms of the sequence in which the $n$-th term is given by:
(i) $t_n = 3n - 4$
(ii) $t_n = 2 - 5n$
(iii) $t_n = n^2 - 2n + 3$ for $n \ge 1$.
(i) $t_n = 3n - 4$
(ii) $t_n = 2 - 5n$
(iii) $t_n = n^2 - 2n + 3$ for $n \ge 1$.
(i) $t_n = 3n - 4$:
• $n = 1 \implies t_1 = 3(1) - 4 = -1$
• $n = 2 \implies t_2 = 3(2) - 4 = 2$
• $n = 3 \implies t_3 = 3(3) - 4 = 5$
• $n = 4 \implies t_4 = 3(4) - 4 = 8$
• $n = 5 \implies t_5 = 3(5) - 4 = 11$
First five terms: $-1, 2, 5, 8, 11$
• $n = 1 \implies t_1 = 3(1) - 4 = -1$
• $n = 2 \implies t_2 = 3(2) - 4 = 2$
• $n = 3 \implies t_3 = 3(3) - 4 = 5$
• $n = 4 \implies t_4 = 3(4) - 4 = 8$
• $n = 5 \implies t_5 = 3(5) - 4 = 11$
First five terms: $-1, 2, 5, 8, 11$
(ii) $t_n = 2 - 5n$:
• $n = 1 \implies t_1 = 2 - 5(1) = -3$
• $n = 2 \implies t_2 = 2 - 5(2) = -8$
• $n = 3 \implies t_3 = 2 - 5(3) = -13$
• $n = 4 \implies t_4 = 2 - 5(4) = -18$
• $n = 5 \implies t_5 = 2 - 5(5) = -23$
First five terms: $-3, -8, -13, -18, -23$
• $n = 1 \implies t_1 = 2 - 5(1) = -3$
• $n = 2 \implies t_2 = 2 - 5(2) = -8$
• $n = 3 \implies t_3 = 2 - 5(3) = -13$
• $n = 4 \implies t_4 = 2 - 5(4) = -18$
• $n = 5 \implies t_5 = 2 - 5(5) = -23$
First five terms: $-3, -8, -13, -18, -23$
(iii) $t_n = n^2 - 2n + 3$:
• $n = 1 \implies t_1 = 1^2 - 2(1) + 3 = 1 - 2 + 3 = 2$
• $n = 2 \implies t_2 = 2^2 - 2(2) + 3 = 4 - 4 + 3 = 3$
• $n = 3 \implies t_3 = 3^2 - 2(3) + 3 = 9 - 6 + 3 = 6$
• $n = 4 \implies t_4 = 4^2 - 2(4) + 3 = 16 - 8 + 3 = 11$
• $n = 5 \implies t_5 = 5^2 - 2(5) + 3 = 25 - 10 + 3 = 18$
First five terms: $2, 3, 6, 11, 18$
• $n = 1 \implies t_1 = 1^2 - 2(1) + 3 = 1 - 2 + 3 = 2$
• $n = 2 \implies t_2 = 2^2 - 2(2) + 3 = 4 - 4 + 3 = 3$
• $n = 3 \implies t_3 = 3^2 - 2(3) + 3 = 9 - 6 + 3 = 6$
• $n = 4 \implies t_4 = 4^2 - 2(4) + 3 = 16 - 8 + 3 = 11$
• $n = 5 \implies t_5 = 5^2 - 2(5) + 3 = 25 - 10 + 3 = 18$
First five terms: $2, 3, 6, 11, 18$
(i) $-1, 2, 5, 8, 11$
(ii) $-3, -8, -13, -18, -23$
(iii) $2, 3, 6, 11, 18$
Q2: Specific Terms Calculation
Find the 10th and 15th terms of the sequence $t_n = 5n - 3$ for $n \ge 1$.
10th Term ($n = 10$):
$$t_{10} = 5(10) - 3 = 50 - 3 = \mathbf{47}$$
$$t_{10} = 5(10) - 3 = 50 - 3 = \mathbf{47}$$
15th Term ($n = 15$):
$$t_{15} = 5(15) - 3 = 75 - 3 = \mathbf{72}$$
$$t_{15} = 5(15) - 3 = 75 - 3 = \mathbf{72}$$
10th term = 47, 15th term = 72
Q3: Value Membership Check
Determine whether $97$ and $172$ are terms of the sequence $t_n = 5n - 3$ for $n \ge 1$.
Check for 97:
Set $5n - 3 = 97$:
$$5n = 97 + 3 \implies 5n = 100 \implies n = \frac{100}{5} = 20$$
Since $20$ is a natural number, 97 is the 20th term of the sequence.
Set $5n - 3 = 97$:
$$5n = 97 + 3 \implies 5n = 100 \implies n = \frac{100}{5} = 20$$
Since $20$ is a natural number, 97 is the 20th term of the sequence.
Check for 172:
Set $5n - 3 = 172$:
$$5n = 172 + 3 \implies 5n = 175 \implies n = \frac{175}{5} = 35$$
Since $35$ is a natural number, 172 is the 35th term of the sequence.
Set $5n - 3 = 172$:
$$5n = 172 + 3 \implies 5n = 175 \implies n = \frac{175}{5} = 35$$
Since $35$ is a natural number, 172 is the 35th term of the sequence.
Both 97 (20th term) and 172 (35th term) are terms of the sequence.
Q4: Find Position of Term
Which term of the sequence $t_n = 5n - 3$ for $n \ge 1$ is $607$?
Set $t_n = 607$ and solve for $n$:
$$5n - 3 = 607$$
$$5n = 607 + 3$$
$$5n = 610$$
$$n = \frac{610}{5} = \mathbf{122}$$
$$5n - 3 = 607$$
$$5n = 607 + 3$$
$$5n = 610$$
$$n = \frac{610}{5} = \mathbf{122}$$
Since $122$ is a natural number, $607$ is the **122nd term** of the sequence.
607 is the 122nd term
Q5: Recursive Rule & Membership
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?
Find the first five terms:
• $t_1 = -5$
• $t_2 = t_1 + 3 = -5 + 3 = -2$
• $t_3 = t_2 + 3 = -2 + 3 = 1$
• $t_4 = t_3 + 3 = 1 + 3 = 4$
• $t_5 = t_4 + 3 = 4 + 3 = 7$
First five terms: $-5, -2, 1, 4, 7$
• $t_1 = -5$
• $t_2 = t_1 + 3 = -5 + 3 = -2$
• $t_3 = t_2 + 3 = -2 + 3 = 1$
• $t_4 = t_3 + 3 = 1 + 3 = 4$
• $t_5 = t_4 + 3 = 4 + 3 = 7$
First five terms: $-5, -2, 1, 4, 7$
Check if 52 is a term:
Observe the sequence is an Arithmetic Progression (AP) with first term $a = -5$ and common difference $d = 3$.
The explicit formula for the $n$-th term is:
$$t_n = a + (n - 1)d = -5 + (n - 1)3 = -5 + 3n - 3 = 3n - 8$$
Set $3n - 8 = 52$ to check:
$$3n = 52 + 8 \implies 3n = 60 \implies n = 20$$
Since $20$ is a natural number, **52 is a term**, specifically the **20th term** of the sequence.
Observe the sequence is an Arithmetic Progression (AP) with first term $a = -5$ and common difference $d = 3$.
The explicit formula for the $n$-th term is:
$$t_n = a + (n - 1)d = -5 + (n - 1)3 = -5 + 3n - 3 = 3n - 8$$
Set $3n - 8 = 52$ to check:
$$3n = 52 + 8 \implies 3n = 60 \implies n = 20$$
Since $20$ is a natural number, **52 is a term**, specifically the **20th term** of the sequence.
First 5 terms: $-5, -2, 1, 4, 7$; 52 is the 20th term.
Q6: Higher-Order Recursive Relation (Tribonacci)
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 the terms $T_4, T_5, T_6, T_7,$ and $T_8$.
Find the terms $T_4, T_5, T_6, T_7,$ and $T_8$.
Calculate each term sequentially by summing the preceding three terms:
• $T_4 = T_3 + T_2 + T_1 = 4 + 2 + 1 = \mathbf{7}$
• $T_5 = T_4 + T_3 + T_2 = 7 + 4 + 2 = \mathbf{13}$
• $T_6 = T_5 + T_4 + T_3 = 13 + 7 + 4 = \mathbf{24}$
• $T_7 = T_6 + T_5 + T_4 = 24 + 13 + 7 = \mathbf{44}$
• $T_8 = T_7 + T_6 + T_5 = 44 + 24 + 13 = \mathbf{81}$
• $T_4 = T_3 + T_2 + T_1 = 4 + 2 + 1 = \mathbf{7}$
• $T_5 = T_4 + T_3 + T_2 = 7 + 4 + 2 = \mathbf{13}$
• $T_6 = T_5 + T_4 + T_3 = 13 + 7 + 4 = \mathbf{24}$
• $T_7 = T_6 + T_5 + T_4 = 24 + 13 + 7 = \mathbf{44}$
• $T_8 = T_7 + T_6 + T_5 = 44 + 24 + 13 = \mathbf{81}$
T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81