Chapter 8

Chapter 8: Predicting What Comes Next — Solutions Guide

1. Section 8.1 & 8.2: In-Text Exercises

Exercise 8.1 (Page 177): Consider the sequence 1, 4, 7, 10, 13...
(i) Can you predict the next four terms?
(ii) Can you derive the first 10 terms of the sequence obtained by adding all the terms up to a given term of this sequence?

Solution:

(i) Next four terms:
The pattern shows adding 3 to the previous term ($t_n = t_{n-1} + 3$).
Next four terms: 13 + 3 = 16, 16 + 3 = 19, 19 + 3 = 22, 22 + 3 = 25.
Next 4 terms: 16, 19, 22, 25

(ii) Sequence of partial sums (first 10 terms):
Let $S_k$ be the sum of the first $k$ terms:

• $S_1 = 1$
• $S_2 = 1 + 4 = 5$
• $S_3 = 5 + 7 = 12$
• $S_4 = 12 + 10 = 22$
• $S_5 = 22 + 13 = 35$
• $S_6 = 35 + 16 = 51$
• $S_7 = 51 + 19 = 70$
• $S_8 = 70 + 22 = 92$
• $S_9 = 92 + 25 = 117$
• $S_{10} = 117 + 28 = 145$

First 10 sum terms: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145

Exercise 8.2 (Page 178): Using the explicit rule $u_n = 2n - 1$, find the 53rd term, the 108th term, and the 1170th term of the odd number sequence.

Solution:

• For $n = 53$: $u_{53} = 2(53) - 1 = 106 - 1 = 105$
• For $n = 108$: $u_{108} = 2(108) - 1 = 216 - 1 = 215$
• For $n = 1170$: $u_{1170} = 2(1170) - 1 = 2340 - 1 = 2339$

53rd term = 105, 108th term = 215, 1170th term = 2339

Exercise 8.3 (Page 179): Consider the expression $t_n = 3n - 7$.
(i) Find its 1st, 2nd, 3rd, 12th, 18th, and 50th terms.
(ii) Which term of the sequence is 332?
(iii) Is 557 a term of this sequence? Why or why not?

Solution:

(i) Finding terms:

• $t_1 = 3(1) - 7 = -4$
• $t_2 = 3(2) - 7 = -1$
• $t_3 = 3(3) - 7 = 2$
• $t_{12} = 3(12) - 7 = 36 - 7 = 29$
• $t_{18} = 3(18) - 7 = 54 - 7 = 47$
• $t_{50} = 3(50) - 7 = 150 - 7 = 143$

(ii) Which term is 332?
Set $3n - 7 = 332 \implies 3n = 339 \implies n = 113$.
332 is the 113th term.

(iii) Is 557 a term?
Set $3n - 7 = 557 \implies 3n = 564 \implies n = 188$.
Since 188 is a natural number ($n \in \mathbb{N}$), Yes, 557 is the 188th term.

2. EXERCISE SET 8.1 (Page 180)

Question 1: 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 $n \ge 1$.

Solution:

(i) $t_n = 3n - 4$:
$t_1 = 3(1)-4 = -1$, $t_2 = 3(2)-4 = 2$, $t_3 = 3(3)-4 = 5$, $t_4 = 3(4)-4 = 8$, $t_5 = 3(5)-4 = 11$.
-1, 2, 5, 8, 11

(ii) $t_n = 2 - 5n$:
$t_1 = 2-5(1) = -3$, $t_2 = 2-5(2) = -8$, $t_3 = 2-5(3) = -13$, $t_4 = 2-5(4) = -18$, $t_5 = 2-5(5) = -23$.
-3, -8, -13, -18, -23

(iii) $t_n = n^2 - 2n + 3$:
$t_1 = 1^2 - 2(1) + 3 = 2$
$t_2 = 2^2 - 2(2) + 3 = 3$
$t_3 = 3^2 - 2(3) + 3 = 6$
$t_4 = 4^2 - 2(4) + 3 = 11$
$t_5 = 5^2 - 2(5) + 3 = 18$
2, 3, 6, 11, 18

Question 2: Find the 10th and 15th terms of the sequence $t_n = 5n - 3$ for $n \ge 1$.

• $t_{10} = 5(10) - 3 = 50 - 3 = 47$
• $t_{15} = 5(15) - 3 = 75 - 3 = 72$

10th term = 47, 15th term = 72

Question 3: Determine whether 97 and 172 are terms of the sequence $t_n = 5n - 3$ for $n \ge 1$.

For 97: $5n - 3 = 97 \implies 5n = 100 \implies n = 20$. Since 20 is a natural number, 97 is a term (20th term).

For 172: $5n - 3 = 172 \implies 5n = 175 \implies n = 35$. Since 35 is a natural number, 172 is a term (35th term).

Both 97 (20th term) and 172 (35th term) are terms of the sequence.

Question 4: Which term of the sequence $t_n = 5n - 3$ for $n \ge 1$ is 607?

$5n - 3 = 607 \implies 5n = 610 \implies n = 122$.

607 is the 122nd term.

Question 5: 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?

First five terms:

• $t_1 = -5$
• $t_2 = t_1 + 3 = -5 + 3 = -2$
• $t_3 = -2 + 3 = 1$
• $t_4 = 1 + 3 = 4$
• $t_5 = 4 + 3 = 7$

Explicit Rule Derivation: $t_n = a + (n-1)d = -5 + (n-1)3 = 3n - 8$.

Check for 52: $3n - 8 = 52 \implies 3n = 60 \implies n = 20$.

First five terms: -5, -2, 1, 4, 7. Yes, 52 is the 20th term.

Question 6: 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 $T_8$.

• $T_4 = T_3 + T_2 + T_1 = 4 + 2 + 1 = 7$
• $T_5 = T_4 + T_3 + T_2 = 7 + 4 + 2 = 13$
• $T_6 = T_5 + T_4 + T_3 = 13 + 7 + 4 = 24$
• $T_7 = T_6 + T_5 + T_4 = 24 + 13 + 7 = 44$
• $T_8 = T_7 + T_6 + T_5 = 44 + 24 + 13 = 81$

T₄=7, T₅=13, T₆=24, T₇=44, T₈=81

3. Section 8.5 Exercise Set (Page 186)

Question 1: Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ....

First term $a = 3$, common difference $d = 8 - 3 = 5$. Formula: $t_n = a + (n-1)d$.

• $t_{10} = 3 + (10-1)5 = 3 + 45 = 48$
• $t_{26} = 3 + (26-1)5 = 3 + 125 = 128$

10th term = 48, 26th term = 128

Question 2: Which term of the AP: 21, 18, 15, ... is -81? Also, is 0 a term of this AP? Give reasons for your answer.

Here, $a = 21$, $d = 18 - 21 = -3$. Formula: $t_n = 21 + (n-1)(-3) = 24 - 3n$.

For -81: $24 - 3n = -81 \implies 3n = 105 \implies n = 35$. Thus, -81 is the 35th term.

For 0: $24 - 3n = 0 \implies 3n = 24 \implies n = 8$. Since 8 is a natural number, 0 is the 8th term.

-81 is the 35th term. Yes, 0 is a term (the 8th term).

Question 3: Find the $n$-th term of the AP: 11, 8, 5, 2... Write the recursive rule for this AP.

Here $a = 11$, $d = 8 - 11 = -3$.

Explicit Rule ($n$-th term): $t_n = 11 + (n-1)(-3) = 14 - 3n$.

Recursive Rule: $t_1 = 11$, $t_n = t_{n-1} - 3$ for $n \ge 2$.

n-th term: 14 - 3n | Recursive rule: t₁ = 11, tₙ = tₙ₋₁ - 3 (n ≥ 2)

Question 4: An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.

Given $t_3 = a + 2d = 12$ and $t_{50} = a + 49d = 106$.

Subtracting the first equation from the second:
$(a + 49d) - (a + 2d) = 106 - 12 \implies 47d = 94 \implies d = 2$.

Substitute $d = 2$ into $a + 2d = 12 \implies a + 4 = 12 \implies a = 8$.

Finding $t_{29}$: $t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64$.

29th term = 64

Question 5: How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

2-digit numbers divisible by 3 form an AP: 12, 15, 18, ..., 99.

Here $a = 12, d = 3$, last term $l = 99$.

Find total number of terms $n$:
$99 = 12 + (n-1)3 \implies 87 = 3(n-1) \implies n-1 = 29 \implies n = 30$.

Calculate sum $S_{30}$ using $S_n = \frac{n}{2}(a + l)$:
$S_{30} = \frac{30}{2}(12 + 99) = 15 \times 111 = 1665$.

There are 30 numbers, and their sum is 1665.

Question 6: Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

This forms an AP: $a = 5,00,000$, $d = 20,000$, $t_n = 7,00,000$.

$5,00,000 + (n-1)20,000 = 7,00,000$
$(n-1)20,000 = 2,00,000 \implies n-1 = 10 \implies n = 11$.

Income reaches ₹7,00,000 in the 11th year (after 10 years of service/increments).

After 10 years (in the 11th year).

Question 7: A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?

This requires the sum of the first 25 natural numbers ($n = 25$).

Using $S_n = \frac{n(n+1)}{2}$:
$S_{25} = \frac{25 \times 26}{2} = 25 \times 13 = 325$.

Total marbles used = 325

4. Section 8.6: Geometric Progression Exercises (Page 189)

Exercise (Page 189): Check whether the following sequences are geometric progressions and find their $n$-th terms.
(i) 2, 10, 50, 250, ...
(ii) $4, \frac{8}{3}, \frac{16}{9}, \frac{32}{27}, ...$
(iii) $3, \frac{-3}{2}, \frac{3}{4}, \frac{-3}{8}, ...$

(i) 2, 10, 50, 250, ...
Ratios: $\frac{10}{2} = 5, \frac{50}{10} = 5, \frac{250}{50} = 5$. Common ratio $r = 5$, $a = 2$.
It is a GP. $n$-th term $t_n = 2 \times 5^{n-1}$

(ii) $4, \frac{8}{3}, \frac{16}{9}, \frac{32}{27}, ...$
Ratios: $\frac{8/3}{4} = \frac{2}{3}, \frac{16/9}{8/3} = \frac{2}{3}$. Common ratio $r = \frac{2}{3}$, $a = 4$.
It is a GP. $n$-th term $t_n = 4 \times \left(\frac{2}{3}\right)^{n-1}$

(iii) $3, \frac{-3}{2}, \frac{3}{4}, \frac{-3}{8}, ...$
Ratios: $\frac{-3/2}{3} = -\frac{1}{2}, \frac{3/4}{-3/2} = -\frac{1}{2}$. Common ratio $r = -\frac{1}{2}$, $a = 3$.
It is a GP. $n$-th term $t_n = 3 \times \left(-\frac{1}{2}\right)^{n-1}$

Exercise (Page 189): Can you find a recursive rule for the formula $t_n = 3 \times 10^{n-1}$ that generates the geometric progression 3, 30, 300, 3000, ...?

First term $t_1 = 3$. Each term is obtained by multiplying the preceding term by 10.

Recursive Rule: $t_1 = 3$, $t_n = 10 \cdot t_{n-1}$ for $n \ge 2$.