--- title: "NCERT Solutions Class 9 Maths Predicting What Comes Next: Exploring Sequences and Progressions" url: https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions dateModified: 2026-10-07T15:49:15+00:00 --- # NCERT Solutions Class 9 Maths Predicting What Comes Next: Exploring Sequences and Progressions This chapter's questions cover concepts related to sequences, explicit and recursive rules, arithmetic progressions, geometric progressions, and the sum of natural numbers. Free PDF (21 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-8-predicting-what-comes-next-exploring-sequences-and-progressions-13ea0a2ffd.pdf ## Exercise: Consider the sequence 1, 4, 7, 10, 13, ... ### Question 1 *3 marks · Short answer* Can you predict the next four terms? Can you derive the first 10 terms of the sequence obtained by adding all the terms up to a given term of this sequence? (Hint: The first term is 1. The second term is 1 + 4 = 5, the third term is 1 + 4 + 7 = 12, and so on.) **Solution** 1. The given sequence is $1, 4, 7, 10, 13, \dots$, where each term increases by $3$. 2. The next four terms are $16, 19, 22$ and $25$. 3. The first 10 terms of the sequence obtained by adding all terms up to each position are $1, 5, 12, 22, 35, 51, 70, 92, 117$ and $145$. **Answer:** Next four terms: $16, 19, 22, 25$; First 10 cumulative sums: $1, 5, 12, 22, 35, 51, 70, 92, 117, 145$ > Common mistake: Forgetting to add the previous terms cumulatively when finding the second part of the sequence. ## Exercise: Can you write t5, t6, t7 and t8 for the sequence of triangular numbers? ### Question 1 *2 marks · Very short answer* Can you write $t_5$, $t_6$, $t_7$ and $t_8$ for the sequence of triangular numbers? **Solution** 1. The triangular numbers are given by the formula $t_n = \frac{n(n + 1)}{2}$. 2. Substituting $n = 5, 6, 7, 8$ gives $t_5 = 15$, $t_6 = 21$, $t_7 = 28$, and $t_8 = 36$. **Answer:** $t_5 = 15, t_6 = 21, t_7 = 28, t_8 = 36$ > Common mistake: Confusing triangular numbers with square numbers or natural numbers. ## Exercise: Using the explicit rule un = 2n – 1, find the 53rd term, the 108th term, and the 1170th term of the odd number sequence. ### Question 1 *3 marks · Short answer* Using the explicit rule $u_n = 2n - 1$, find the $53^{\text{rd}}$ term, the $108^{\text{th}}$ term, and the $1170^{\text{th}}$ term of the odd number sequence. **Part (i)** 1. Substitute $n = 53$ in the explicit rule $u_n = 2n - 1$. 2. $u_{53} = 2(53) - 1 = 106 - 1 = 105$ Answer (i): 105 **Part (ii)** 1. Substitute $n = 108$ in the explicit rule $u_n = 2n - 1$. 2. $u_{108} = 2(108) - 1 = 216 - 1 = 215$ Answer (ii): 215 **Part (iii)** 1. Substitute $n = 1170$ in the explicit rule $u_n = 2n - 1$. 2. $u_{1170} = 2(1170) - 1 = 2340 - 1 = 2339$ Answer (iii): 2339 **Answer:** The 53rd, 108th, and 1170th terms are 105, 215, and 2339 respectively. > Common mistake: Forgetting to subtract 1 after multiplying the term position by 2. ## Exercise: Consider the expression tn = 3n – 7. ### Question 1 *3 marks · Short answer* (i) Find its first, second, third, $12^{\text{th}}$, $18^{\text{th}}$ and $50^{\text{th}}$ terms. (ii) Which term of the sequence is 332? (iii) Is 557 a term of this sequence? Why or why not? **Part (i)** 1. Given the explicit rule $t_n = 3n - 7$. 2. Substitute $n = 1, 2, 3, 12, 18, 50$ to find the respective terms. 3. $t_1 = 3(1) - 7 = -4$, $t_2 = 3(2) - 7 = -1$, and $t_3 = 3(3) - 7 = 2$. 4. $t_{12} = 3(12) - 7 = 29$, $t_{18} = 3(18) - 7 = 47$, and $t_{50} = 3(50) - 7 = 143$. Answer (i): -4, -1, 2, 29, 47, 143 **Part (ii)** 1. Set $t_n = 332$ in the explicit formula $3n - 7 = 332$. 2. Solve for $n$: $3n = 332 + 7 = 339$. 3. $n = \frac{339}{3} = 113$. 4. Thus, 332 is the 113th term of the sequence. Answer (ii): 113th term **Part (iii)** 1. Assume 557 is a term, so $3n - 7 = 557$. 2. Solve for $n$: $3n = 557 + 7 = 564$. 3. $n = \frac{564}{3} = 188$. 4. Since 188 is a natural number, 557 is a term of this sequence. Answer (iii): Yes, 557 is a term (the 188th term) **Answer:** First term is -4, second is -1, third is 2, 12th is 29, 18th is 47, 50th is 143; 332 is the 114th term; 557 is not a term because n is not a natural number. > Common mistake: Forgetting that $n$ must be a natural number to be a term position in a sequence. ## Exercise Set 8.1 ### Question 1 *3 marks · Short answer* Find the first five terms of the sequence in which the $n^{\text{th}}$ term is given by (i) $t_n = 3n - 4$, (ii) $t_n = 2 - 5n$, and (iii) $t_n = n^2 - 2n + 3$ for $n \geq 1$. **Part (i)** 1. For $t_n = 3n - 4$, substitute $n = 1, 2, 3, 4, 5$. 2. $t_1 = 3(1) - 4 = -1$, $t_2 = 3(2) - 4 = 2$, $t_3 = 3(3) - 4 = 5$. 3. $t_4 = 3(4) - 4 = 8$, $t_5 = 3(5) - 4 = 11$. Answer (i): -1, 2, 5, 8, 11 **Part (ii)** 1. For $t_n = 2 - 5n$, substitute $n = 1, 2, 3, 4, 5$. 2. $t_1 = 2 - 5(1) = -3$, $t_2 = 2 - 5(2) = -8$, $t_3 = 2 - 5(3) = -13$. 3. $t_4 = 2 - 5(4) = -18$, $t_5 = 2 - 5(5) = -23$. Answer (ii): -3, -8, -13, -18, -23 **Part (iii)** 1. For $t_n = n^2 - 2n + 3$, substitute $n = 1, 2, 3, 4, 5$. 2. $t_1 = 1^2 - 2(1) + 3 = 2$, $t_2 = 2^2 - 2(2) + 3 = 3$, $t_3 = 3^2 - 2(3) + 3 = 6$. 3. $t_4 = 4^2 - 2(4) + 3 = 11$, $t_5 = 5^2 - 2(5) + 3 = 18$. Answer (iii): 2, 3, 6, 11, 18 **Answer:** (i) -1, 2, 5, 8, 11; (ii) -3, -8, -13, -18, -23; (iii) 2, 3, 6, 11, 18 > Common mistake: Arithmetic errors when substituting negative values or powers. ### Question 2 *2 marks · Very short answer* Find the $10^{\text{th}}$ and $15^{\text{th}}$ terms of the sequence $t_n = 5n - 3$ for $n \geq 1$. **Solution** 1. The given explicit formula is $t_n = 5n - 3$. 2. Substitute $n = 10$ to get $t_{10} = 5(10) - 3 = 47$. 3. Substitute $n = 15$ to get $t_{15} = 5(15) - 3 = 72$. **Answer:** The $10^{\text{th}}$ term is $47$ and the $15^{\text{th}}$ term is $72$. > Common mistake: Multiplying incorrectly or making errors in subtraction. ### Question 3 *3 marks · Short answer* Determine whether 97 and 172 are terms of the sequence $t_n = 5n - 3$ for $n \geq 1$. **Solution** 1. Equate the given explicit formula $t_n = 5n - 3$ to 97 to check if 97 is a term. 2. $5n - 3 = 97 \implies 5n = 100 \implies n = 20$. Since 20 is a natural number, 97 is a term. 3. Equate $5n - 3$ to 172 to check if 172 is a term. 4. $5n - 3 = 172 \implies 5n = 175 \implies n = 35$. Since 35 is a natural number, 172 is a term. **Answer:** Both 97 and 172 are terms of the sequence (specifically the $20^{\text{th}}$ and $35^{\text{th}}$ terms respectively). > Common mistake: Assuming a number is not a term without fully solving for $n$ as a natural number. ### Question 4 *2 marks · Very short answer* Which term of the sequence $t_n = 5n - 3$ for $n \geq 1$ is 607? **Solution** 1. Set $t_n = 607$ using the formula $t_n = 5n - 3$. 2. $5n - 3 = 607 \implies 5n = 610 \implies n = 122$. **Answer:** 607 is the $122^{\text{nd}}$ term of the sequence. > Common mistake: Calculation error while dividing 610 by 5. ### Question 5 *3 marks · Short answer* A sequence is given by the recursive rule $t_1 = -5$, $t_{n+1} = t_n + 3$ for $n \geq 1$. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it? **Solution** 1. Given $t_1 = -5$ and $t_{n+1} = t_n + 3$. 2. $t_2 = t_1 + 3 = -5 + 3 = -2$, $t_3 = t_2 + 3 = -2 + 3 = 1$, $t_4 = t_3 + 3 = 1 + 3 = 4$, $t_5 = t_4 + 3 = 4 + 3 = 7$. 3. To check if 52 is a term, we observe that the sequence is an AP with $a = -5$ and $d = 3$, so its explicit rule is $t_n = a + (n-1)d = -5 + (n-1)3 = 3n - 8$. 4. Set $3n - 8 = 52 \implies 3n = 60 \implies n = 20$. Since 20 is a natural number, 52 is a term. **Answer:** The first five terms are $-5, -2, 1, 4, 7$. Yes, 52 is a term, and it is the $20^{\text{th}}$ term. > Common mistake: Mistaking the recursive formula to find the position directly without finding the explicit formula or stepping through. ### Question 6 *3 marks · Short answer* Let $T_1 = 1$, $T_2 = 2$, $T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \geq 4$. Find $T_4$, $T_5$, $T_6$, $T_7$, and $T_8$. **Solution** 1. Given $T_1 = 1$, $T_2 = 2$, $T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \geq 4$. 2. $T_4 = T_3 + T_2 + T_1 = 4 + 2 + 1 = 7$. 3. $T_5 = T_4 + T_3 + T_2 = 7 + 4 + 2 = 13$, $T_6 = T_5 + T_4 + T_3 = 13 + 7 + 4 = 24$. 4. $T_7 = T_6 + T_5 + T_4 = 24 + 13 + 7 = 44$, $T_8 = T_7 + T_6 + T_5 = 44 + 24 + 13 = 81$. **Answer:** $T_4 = 7$, $T_5 = 13$, $T_6 = 24$, $T_7 = 44$, $T_8 = 81$. > Common mistake: Adding wrong previous terms due to indexing confusion. ## Exercise: Verify that the following sequences are arithmetic progressions and write their nth terms. ### Question 1 *3 marks · Short answer* Verify that the following sequences are arithmetic progressions and write their $n^{\text{th}}$ terms. What do you observe when you plot the ordered pairs emerging from them? (i) 2, 5, 8, 11, ... (ii) $-5, -1, 3, 7, ...$ **Part (i)** 1. The given sequence is $2, 5, 8, 11, \dots$. 2. The difference between consecutive terms is $5 - 2 = 3$, $8 - 5 = 3$, and $11 - 8 = 3$, which is a constant common difference $d = 3$. 3. Using $t_n = a + (n - 1)d$ with $a = 2$ and $d = 3$, we get $t_n = 2 + (n - 1) \times 3 = 3n - 1$. 4. When the ordered pairs $(x, y)$ representing stage number and term value are plotted, they lie on a straight line. Answer (i): AP verified with common difference $3$, $n^{\text{th}}$ term $t_n = 3n - 1$, and points lie on a straight line. **Part (ii)** 1. The given sequence is $-5, -1, 3, 7, \dots$. 2. The difference between consecutive terms is $-1 - (-5) = 4$, $3 - (-1) = 4$, and $7 - 3 = 4$, which is a constant common difference $d = 4$. 3. Using $t_n = a + (n - 1)d$ with $a = -5$ and $d = 4$, we get $t_n = -5 + (n - 1) \times 4 = 4n - 9$. 4. When the ordered pairs $(x, y)$ representing stage number and term value are plotted, they lie on a straight line. Answer (ii): AP verified with common difference $4$, $n^{\text{th}}$ term $t_n = 4n - 9$, and points lie on a straight line. **Answer:** For sequence (i), $t_n = 3n - 1$; for sequence (ii), $t_n = 4n - 9$. Both sets of ordered pairs lie on a straight line. > Common mistake: Mixing up signs when finding the common difference for negative terms in an AP. ## Exercise: Using the formula tn = a + (n – 1) × d, find the nth term of the following arithmetic progressions. ### Question 1 *3 marks · Short answer* Using the formula $t_n = a + (n - 1) \times d$, find the $n^{\text{th}}$ term of the following arithmetic progressions. (i) $\frac{1}{2}, \frac{5}{2}, \frac{9}{2}, \frac{13}{2}, ...$ (ii) $1.5, 3.5, 5.5, 7.5, ...$ **Part (i)** 1. Here, the first term $a = \frac{1}{2}$ and the common difference $d = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2$. 2. Substitute $a = \frac{1}{2}$ and $d = 2$ in the formula $t_n = a + (n - 1) \times d$. 3. $t_n = \frac{1}{2} + (n - 1) \times 2 = \frac{1}{2} + 2n - 2 = 2n - \frac{3}{2}$. Answer (i): $2n - \frac{3}{2}$ **Part (ii)** 1. Here, the first term $a = 1.5$ and the common difference $d = 3.5 - 1.5 = 2$. 2. Substitute $a = 1.5$ and $d = 2$ in the formula $t_n = a + (n - 1) \times d$. 3. $t_n = 1.5 + (n - 1) \times 2 = 1.5 + 2n - 2 = 2n - 0.5$. Answer (ii): $2n - 0.5$ **Answer:** (i) $2n - \frac{3}{2}$, (ii) $2n - 0.5$ > Common mistake: Subtracting terms in the wrong order while finding the common difference $d$. ## Exercise Set 8.2 ### Question 1 *2 marks · Very short answer* Find the $10^{\text{th}}$ and $26^{\text{th}}$ terms of the AP: 3, 8, 13, 18, ... **Solution** 1. Given AP is 3, 8, 13, 18, ..., with first term $a = 3$ and common difference $d = 8 - 3 = 5$. 2. Using the formula $t_n = a + (n - 1)d$, the 10th term is $t_{10} = 3 + (10 - 1) \times 5 = 3 + 45 = 48$ and the 26th term is $t_{26} = 3 + (26 - 1) \times 5 = 3 + 125 = 128$. **Answer:** The 10th term is 48 and the 26th term is 128. > Common mistake: Using $n$ instead of $n-1$ in the formula. ### Question 2 *3 marks · Short answer* Which term of the AP: 21, 18, 15, ... is $-81$? Also, is 0 a term of this AP? Give reasons for your answer. **Solution** 1. Given AP is 21, 18, 15, ..., with $a = 21$ and $d = 18 - 21 = -3$. 2. To find which term is $-81$, we set $t_n = -81$, giving $21 + (n - 1)(-3) = -81$, which simplifies to $-3(n - 1) = -102$, so $n - 1 = 34$ and $n = 35$. Thus, $-81$ is the 35th term. 3. To check if 0 is a term, we set $t_n = 0$, giving $21 + (n - 1)(-3) = 0$, which leads to $3(n - 1) = 21$, so $n - 1 = 7$ and $n = 8$. Since $n$ is a natural number, 0 is the 8th term of this AP. **Answer:** -81 is the 35th term, and 0 is the 8th term of the AP. > Common mistake: Treating non-integer values of $n$ as valid term positions. ### Question 3 *3 marks · Short answer* Find the $n^{\text{th}}$ term of the AP: 11, 8, 5, 2, ... Write the recursive rule for this AP. **Solution** 1. Given AP is 11, 8, 5, 2, ..., with first term $a = 11$ and common difference $d = 8 - 11 = -3$. 2. The $n^{\text{th}}$ term is given by $t_n = a + (n - 1)d = 11 + (n - 1)(-3) = 11 - 3n + 3 = 14 - 3n$. 3. The recursive rule for this AP is $t_1 = 11$ and $t_n = t_{n - 1} - 3$ for $n \ge 2$. **Answer:** The $n^{\text{th}}$ term is $14 - 3n$ and the recursive rule is $t_1 = 11, t_n = t_{n-1} - 3$ for $n \ge 2$. > Common mistake: Incorrect sign when finding the common difference $d$. ### Question 4 *3 marks · Short answer* An AP consists of 50 terms in which the $3^{\text{rd}}$ term is 12 and the last term is 106. Find the $29^{\text{th}}$ term. (Hint: If '$a$' is the first term and '$d$' the common difference, then we arrive at the equations $a + 2d = 12$ and $a + 49d = 106$. Solve this pair of linear equations for '$a$' and '$d$'.) **Solution** 1. Let $a$ be the first term and $d$ be the common difference. Given that the 3rd term is 12, we have $a + 2d = 12$. 2. The last term of 50 terms is 106, so the 50th term is $a + 49d = 106$. 3. Subtracting the first equation from the second gives $47d = 94$, so $d = 2$. Substituting $d = 2$ gives $a = 12 - 4 = 8$. 4. The 29th term is $t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64$. **Answer:** The 29th term of the AP is 64. > Common mistake: Arithmetic errors while solving simultaneous equations. ### Question 5 *3 marks · Short answer* How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers? **Solution** 1. The two-digit numbers divisible by 3 form the AP: 12, 15, 18, ..., 99, with $a = 12$, $d = 3$, and $t_n = 99$. 2. Using $t_n = a + (n - 1)d$, we get $12 + (n - 1)3 = 99$, which gives $3(n - 1) = 87$, so $n - 1 = 29$ and $n = 30$ numbers. 3. The sum of these $n$ numbers can be found using the formula $S_n = \frac{n}{2}(a + l) = \frac{30}{2}(12 + 99) = 15 \times 111 = 1665$. **Answer:** There are 30 such 2-digit numbers and their sum is 1665. > Common mistake: Incorrect identification of the first and last two-digit multiples of 3. ### Question 6 *3 marks · Short answer* 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? **Solution** 1. The annual salaries form an AP with first term $a = 5,00,000$ and common difference $d = 20,000$. 2. The explicit formula for the $n^{\text{th}}$ term is given by $t_n = a + (n - 1)d$. 3. Substituting the values, we get $7,00,000 = 5,00,000 + (n - 1) \times 20,000$. 4. Solving for $n$, we get $2,00,000 = (n - 1) \times 20,000$, which gives $n - 1 = 10$, so $n = 11$. 5. Thus, his income reached $7,00,000$ in the 11th year, which is after 10 years of receiving increments. **Answer:** After 10 years (in the 11th year) > Common mistake: Students often write 11 years instead of 10 years of service, confusing the term number $n = 11$ with the number of elapsed years. ### Question 7 *3 marks · Short answer* 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? **Solution** 1. The number of marbles in each row forms the sequence $1, 2, 3, \dots, 25$, which is the sequence of natural numbers. 2. The total number of marbles used is the sum of the first 25 natural numbers, given by the formula $S_n = \frac{n(n + 1)}{2}$. 3. Substituting $n = 25$, we get $S_{25} = \frac{25(25 + 1)}{2} = \frac{25 \times 26}{2} = 25 \times 13 = 325$. **Answer:** 325 marbles > Common mistake: Multiplying the middle term incorrectly or forgetting to divide by 2 in the sum formula. ## Exercise: Check whether the following sequences are geometric progressions and find their nth terms. ### Question 1 *3 marks · Short answer* Check whether the following sequences are geometric progressions and find their $n^{\text{th}}$ terms. (i) 2, 10, 50, 250, ... (ii) $4, \frac{8}{3}, \frac{16}{9}, \frac{32}{27}, ...$ (iii) $3, -\frac{3}{2}, \frac{3}{4}, -\frac{3}{8}, ...$ **Part (i)** 1. Given sequence: $2, 10, 50, 250, \dots$ 2. Evaluate the ratios of consecutive terms: $\frac{10}{2} = 5$, $\frac{50}{10} = 5$, $\frac{250}{50} = 5$. 3. Since the common ratio $r = 5$ is constant, the given sequence is a geometric progression with first term $a = 2$. 4. The $n^{\text{th}}$ term is given by $t_n = a r^{n-1} = 2 \times 5^{n-1}$. Answer (i): GP, $t_n = 2 \times 5^{n-1}$ **Part (ii)** 1. Given sequence: $4, \frac{8}{3}, \frac{16}{9}, \frac{32}{27}, \dots$ 2. Evaluate the ratios of consecutive terms: $\frac{8/3}{4} = \frac{2}{3}$, $\frac{16/9}{8/3} = \frac{2}{3}$, $\frac{32/27}{16/9} = \frac{2}{3}$. 3. Since the common ratio $r = \frac{2}{3}$ is constant, the sequence is a geometric progression with first term $a = 4$. 4. The $n^{\text{th}}$ term is given by $t_n = a r^{n-1} = 4 \times \left(\frac{2}{3}\right)^{n-1}$. Answer (ii): GP, $t_n = 4 \times \left(\frac{2}{3}\right)^{n-1}$ **Part (iii)** 1. Given sequence: $3, -\frac{3}{2}, \frac{3}{4}, -\frac{3}{8}, \dots$ 2. Evaluate the ratios of consecutive terms: $\frac{-3/2}{3} = -\frac{1}{2}$, $\frac{3/4}{-3/2} = -\frac{1}{2}$, $\frac{-3/8}{3/4} = -\frac{1}{2}$. 3. Since the common ratio $r = -\frac{1}{2}$ is constant, the sequence is a geometric progression with first term $a = 3$. 4. The $n^{\text{th}}$ term is given by $t_n = a r^{n-1} = 3 \times \left(-\frac{1}{2}\right)^{n-1}$. Answer (iii): GP, $t_n = 3 \times \left(-\frac{1}{2}\right)^{n-1}$ **Answer:** The first sequence is a GP with $t_n = 2 \times 5^{n-1}$, the second is a GP with $t_n = 4 \times \left(\frac{2}{3}\right)^{n-1}$, and the third is a GP with $t_n = 3 \times \left(-\frac{1}{2}\right)^{n-1}$. > Common mistake: Dividing the first term by the second term instead of the second term by the first term when finding the common ratio. ## Exercise: Can you find a recursive rule for the formula tn = 3 × 10n–1 that generates the geometric progression 3, 30, 300, 3000, ...? ### Question 1 *2 marks · Very short answer* 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, ...? **Solution** 1. The first term of the sequence is given by $t_1 = 3$. 2. Each term after the first term is obtained by multiplying the previous term by the common ratio $10$, so $t_n = 10 t_{n-1}$ for $n \geq 2$. **Answer:** $t_1 = 3$, $t_n = 10 t_{n-1}$ for $n \geq 2$ > Common mistake: Forgetting to specify the starting term $t_1$ or the condition $n \geq 2$. ## Exercise Set 8.3 ### Question 1 *2 marks · Very short answer* Find the $12^{\text{th}}$ term of a GP with common ratio 2, whose $8^{\text{th}}$ term is 192. **Solution** 1. The $n^{\text{th}}$ term of a GP is given by $t_n = a r^{n-1}$. 2. We are given $t_8 = 192$ and $r = 2$, so $a \times 2^{8-1} = 192 \implies a \times 128 = 192$, which gives $a = \frac{192}{128} = \frac{3}{2}$. 3. The $12^{\text{th}}$ term is $t_{12} = \frac{3}{2} \times 2^{12-1} = \frac{3}{2} \times 2^{11} = 3 \times 2^{10} = 3072$. **Answer:** 3072 > Common mistake: Confusing the exponent in the formula for the $n^{\text{th}}$ term. ### Question 2 *2 marks · Very short answer* Find the $10^{\text{th}}$ and $n^{\text{th}}$ terms of the GP: 5, 25, 125, ... **Solution** 1. The given GP is $5, 25, 125, \dots$, where the first term $a = 5$ and the common ratio $r = \frac{25}{5} = 5$. 2. The $n^{\text{th}}$ term is given by $t_n = a r^{n-1} = 5 \times 5^{n-1} = 5^n$. 3. The $10^{\text{th}}$ term is $t_{10} = 5^{10}$. **Answer:** $t_{10} = 5^{10}$ and $t_n = 5^n$ > Common mistake: Writing $5^{n-1}$ as $25^{n-1}$. ### Question 3 *3 marks · Short answer* A sequence is given by the recursive rule $t_1 = 2$, $t_{n+1} = 3t_n - 2$ for $n \geq 1$. Which term of the sequence is 730? **Solution** 1. The recursive rule is given by $t_1 = 2$ and $t_{n+1} = 3t_n - 2$. 2. Let us find the first few terms: $t_1 = 2$, $t_2 = 3(2) - 2 = 4$, $t_3 = 3(4) - 2 = 10$, $t_4 = 3(10) - 2 = 28$, $t_5 = 3(28) - 2 = 82$, $t_6 = 3(82) - 2 = 244$, $t_7 = 3(244) - 2 = 730$. 3. Thus, 730 is the $7^{\text{th}}$ term of the sequence. **Answer:** 7th term > Common mistake: Arithmetic error while evaluating successive terms using the recursive rule. ### Question 4 *3 marks · Short answer* Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the $n^{\text{th}}$ term. **Solution** 1. The given GP is $2, 6, 18, \dots$, where $a = 2$ and $r = \frac{6}{2} = 3$. 2. The explicit formula for the $n^{\text{th}}$ term is $t_n = 2 \times 3^{n-1}$ and the recursive formula is $t_1 = 2, t_n = 3t_{n-1}$ for $n \geq 2$. 3. To find which term is 4374, substitute $t_n = 4374$ into the explicit formula: $2 \times 3^{n-1} = 4374 \implies 3^{n-1} = 2187$. 4. Since $2187 = 3^7$, we have $n-1 = 7$, which gives $n = 8$. Thus, 4374 is the $8^{\text{th}}$ term. **Answer:** $8^{\text{th}}$ term, explicit formula $t_n = 2 \times 3^{n-1}$, recursive formula $t_1 = 2, t_n = 3t_{n-1}$ > Common mistake: Incorrectly solving the exponential equation for $n$. ### Question 5 *4 marks · Short answer* A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way—each time rising to 60% of the previous height. (i) What height does the ball reach after the $5^{\text{th}}$ bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the $6^{\text{th}}$ time? **Part (i)** 1. Initial height $a = 80\text{ m}$ and common ratio $r = 0.6$. 2. The height after the $n^{\text{th}}$ bounce is given by $t_n = a r^n$. 3. For the $5^{\text{th}}$ bounce, $t_5 = 80 \times (0.6)^5 = 80 \times 0.07776 = 6.2208\text{ m}$. Answer (i): $6.2208\text{ m}$ **Part (ii)** 1. The total vertical distance travelled by the time it hits the ground for the $6^{\text{th}}$ time includes the initial fall of $80\text{ m}$ plus twice the sum of the heights of the first 5 bounces. 2. Distance $= 80 + 2 \times (80 \times 0.6 + 80 \times 0.6^2 + 80 \times 0.6^3 + 80 \times 0.6^4 + 80 \times 0.6^5)$. 3. Calculating the heights: $48, 28.8, 17.28, 10.368, 6.2208$, whose sum is $110.6688\text{ m}$. 4. Total distance $= 80 + 2 \times 110.6688 = 80 + 221.3376 = 301.3376\text{ m}$. Answer (ii): $301.3376\text{ m}$ **Answer:** (i) $6.2208\text{ m}$, (ii) $301.3376\text{ m}$ > Common mistake: Forgetting to multiply the sum of the bounce heights by 2 (since the ball goes up and comes down for every bounce) or forgetting to add the initial drop of 80 m. ### Question 6 *2 marks · Very short answer* Which term of the sequence $2, 2\sqrt{2}, 4, ...$ is 128? **Solution** 1. The sequence is $2, 2\sqrt{2}, 4, \dots$, where the first term $a = 2$ and the common ratio $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$. 2. The $n^{\text{th}}$ term is given by $t_n = a r^{n-1} = 2(\sqrt{2})^{n-1} = 2 \times 2^{\frac{n-1}{2}} = 2^{\frac{n+1}{2}}$. 3. Substitute $t_n = 128$: $2^{\frac{n+1}{2}} = 128 = 2^7$, which gives $\frac{n+1}{2} = 7 \implies n + 1 = 14 \implies n = 13$. **Answer:** $13^{\text{th}}$ term > Common mistake: Errors in handling fractional exponents with $\sqrt{2}$. ### Question 7 *5 marks · Long answer* Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on. Look at Fig. 8.12 and try to answer the following questions. (i) How many red squares are there in Stages 0 to 3? (ii) Can you predict the number of red squares in Stages 4 and 5? (iii) Can you find a rule for the number of red squares at the $n^{\text{th}}$ stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. (iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the $n^{\text{th}}$ stage. What happens to this area as $n$, the number of stages, goes on increasing? **Part (i)** 1. In Stage 0, there is $1$ main square, so $t_0 = 1$. 2. In Stage 1, there are $8$ red squares. 3. In Stage 2, each of the $8$ squares is divided into $8$ smaller ones, giving $8 \times 8 = 64$ squares. 4. In Stage 3, the number of red squares is $64 \times 8 = 512$. 5. Thus, the number of red squares in Stages 0 to 3 are $1$, $8$, $64$, and $512$ respectively. Answer (i): Stages 0 to 3 have 1, 8, 64, and 512 red squares respectively. **Part (ii)** 1. The number of red squares at each stage is multiplied by $8$ to get the next stage. 2. For Stage 4, the number of red squares is $512 \times 8 = 4096$. 3. For Stage 5, the number of red squares is $4096 \times 8 = 32768$. Answer (ii): Stage 4 has 4,096 red squares and Stage 5 has 32,768 red squares. **Part (iii)** 1. The sequence of red squares is $1, 8, 64, 512, \dots$, which is a geometric progression with first term $a = 1$ and common ratio $r = 8$. 2. The explicit formula for the $n^{\text{th}}$ stage is $t_n = 8^n$ for $n \ge 0$. 3. The recursive formula is $t_0 = 1$ and $t_n = 8t_{n-1}$ for $n \ge 1$. Answer (iii): Explicit formula: $t_n = 8^n$; Recursive formula: $t_0 = 1, t_n = 8t_{n-1}$ for $n \ge 1$. **Part (iv)** 1. In Stage 0, the area is $1$. In Stage 1, one out of nine equal parts is removed, leaving $\frac{8}{9}$ of the area. 2. The areas for Stages 1, 2, and 3 are $\frac{8}{9}$, $(\frac{8}{9})^2 = \frac{64}{81}$, and $(\frac{8}{9})^3 = \frac{512}{729}$ square units respectively. 3. For Stage 4 and Stage 5, the areas are $(\frac{8}{9})^4 = \frac{4096}{6561}$ and $(\frac{8}{9})^5 = \frac{32768}{59049}$ square units. 4. The explicit formula for the area at the $n^{\text{th}}$ stage is $s_n = (\frac{8}{9})^n$ and the recursive formula is $s_0 = 1$, $s_n = \frac{8}{9}s_{n-1}$ for $n \ge 1$. 5. As $n$ goes on increasing, the area of the red region decreases and gets closer and closer to $0$. Answer (iv): Areas are $\frac{8}{9}, \frac{64}{81}, \frac{512}{729}, \frac{4096}{6561}, \frac{32768}{59049}$; explicit formula $s_n = (\frac{8}{9})^n$; area approaches $0$ as $n$ increases. **Answer:** The red squares follow $t_n = 8^n$ and the area follows $s_n = (\frac{8}{9})^n$, approaching $0$ as $n$ increases. > Common mistake: Confusing Stage 0 with Stage 1 when writing the starting values for the geometric progression. ## End-of-Chapter Exercises ### Question 1 *3 marks · Short answer* Find the $31^{\text{st}}$ term of an AP whose $11^{\text{th}}$ term is 38 and $16^{\text{th}}$ term is 73. **Solution** 1. Let the first term be $a$ and the common difference be $d$. 2. Given $t_{11} = 38$, so $a + 10d = 38$, and $t_{16} = 73$, so $a + 15d = 73$. 3. Subtracting the first equation from the second gives $5d = 35$, which means $d = 7$. 4. Substituting $d = 7$ into $a + 10d = 38$ gives $a + 70 = 38$, so $a = -32$. 5. The $31^{\text{st}}$ term is $t_{31} = a + 30d = -32 + 30(7) = -32 + 210 = 178$. **Answer:** 178 > Common mistake: Errors in solving simultaneous linear equations for $a$ and $d$. ### Question 2 *3 marks · Short answer* Determine the AP whose third term is 16 and whose $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by 12. **Solution** 1. Let the first term be $a$ and the common difference be $d$. 2. The third term is $t_3 = a + 2d = 16$. 3. The condition that the $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by 12 gives $t_7 - t_5 = 12$, which simplifies to $(a + 6d) - (a + 4d) = 12$, or $2d = 12$, so $d = 6$. 4. Substituting $d = 6$ into $a + 2d = 16$ gives $a + 12 = 16$, so $a = 4$. 5. Thus, the AP is $4, 10, 16, 22, \dots$. **Answer:** $4, 10, 16, 22, \dots$ > Common mistake: Confusing term indices with coefficients. ### Question 3 *3 marks · Short answer* How many three-digit numbers are divisible by 7? (Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.) **Solution** 1. The three-digit numbers divisible by 7 form an AP starting with 105 and ending with 994 with a common difference of 7. 2. The $n^{\text{th}}$ term formula is $t_n = a + (n - 1)d$, which gives $105 + (n - 1)7 = 994$. 3. Solving for $n$ gives $(n - 1)7 = 994 - 105 = 889$. 4. Dividing by 7 gives $n - 1 = 127$, so $n = 128$. 5. There are 128 three-digit numbers divisible by 7. **Answer:** 128 > Common mistake: Incorrectly identifying the smallest or largest three-digit multiple of 7. ### Question 4 *3 marks · Short answer* How many multiples of 4 lie between 10 and 250? (Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.) **Solution** 1. The multiples of 4 lying between 10 and 250 form an AP starting with 12 and ending with 248 with a common difference of 4. 2. Using the $n^{\text{th}}$ term formula $t_n = a + (n - 1)d$, we get $12 + (n - 1)4 = 248$. 3. Subtracting 12 gives $(n - 1)4 = 236$. 4. Dividing by 4 gives $n - 1 = 59$, so $n = 60$. 5. There are 60 multiples of 4 between 10 and 250. **Answer:** 60 > Common mistake: Including numbers outside the range 10 and 250. ### Question 5 *3 marks · Short answer* Find a GP for which the sum of the first two terms is $-4$ and the fifth term is 4 times the third term. **Solution** 1. Let the first term be $a$ and the common ratio be $r$. 2. The sum of the first two terms is $a + ar = -4$, so $a(1 + r) = -4$. 3. The fifth term is 4 times the third term, so $ar^4 = 4ar^2$. 4. Since $a \neq 0$ and $r \neq 0$, dividing by $ar^2$ gives $r^2 = 4$, so $r = 2$ or $r = -2$. 5. If $r = 2$, $a(1 + 2) = -4 \implies a = -\frac{4}{3}$, giving the GP $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$. If $r = -2$, $a(1 - 2) = -4 \implies a = 4$, giving the GP $4, -8, 16, \dots$. **Answer:** $4, -8, 16, \dots \text{ or } -\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$ > Common mistake: Missing one of the possible values for the common ratio $r$. ### Question 6 *3 marks · Short answer* Find all possible ways of expressing 100 as the sum of consecutive natural numbers. **Solution** 1. Let the sum of $k$ consecutive natural numbers starting from $x$ be equal to $100$, where $x \ge 1$ and $k \ge 2$. 2. The sum is given by the formula $\frac{k}{2} [2x + (k - 1)] = 100$, which simplifies to $k(2x + k - 1) = 200$. 3. Here, $k$ and $(2x + k - 1)$ are factors of $200$ having different parities, and $k < 2x + k - 1$. 4. The factor pairs of $200$ with one odd and one even factor are $(1, 200)$, $(8, 25)$, and $(5, 40)$, giving the valid solutions: $100$, $18 + 19 + 20 + 21 + 22$, and $9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$. 5. Thus, the possible ways are $100$, $18 + 19 + 20 + 21 + 22$, and $9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$. **Answer:** 100; 18 + 19 + 20 + 21 + 22; 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 > Common mistake: Forgetting that a single number 100 itself is trivially a sum of one natural number. ### Question 7 *3 marks · Short answer* The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the $2^{\text{nd}}$ hour, $4^{\text{th}}$ hour and $n^{\text{th}}$ hour? **Solution** 1. Initial number of bacteria $a = 30$, and the number doubles every hour, so common ratio $r = 2$. 2. The number of bacteria at the end of the $n^{\text{th}}$ hour is given by the GP formula $t_n = a \cdot r^{n-1}$ or $t_n = 30 \times 2^n$. 3. At the end of the $2^{\text{nd}}$ hour ($n = 2$), the number of bacteria is $30 \times 2^2 = 120$. 4. At the end of the $4^{\text{th}}$ hour ($n = 4$), the number of bacteria is $30 \times 2^4 = 480$. 5. At the end of the $n^{\text{th}}$ hour, the number of bacteria is $30 \times 2^n$. **Answer:** 120 bacteria at the end of the 2nd hour, 480 bacteria at the end of the 4th hour, and $30 \times 2^n$ bacteria at the end of the nth hour. > Common mistake: Using $2^{n-1}$ instead of $2^n$ when counting bacteria at the end of $n$ hours. ### Question 8 *3 marks · Short answer* The sum of the $4^{\text{th}}$ and $8^{\text{th}}$ terms of an AP is 24 and the sum of the $6^{\text{th}}$ and $10^{\text{th}}$ terms is 44. Find the first three terms of the AP. **Solution** 1. Let the first term of the AP be $a$ and the common difference be $d$. 2. The $n^{\text{th}}$ term is given by $t_n = a + (n-1)d$. 3. Given that $t_4 + t_8 = 24$, so $(a + 3d) + (a + 7d) = 24$, which gives $2a + 10d = 24$ or $a + 5d = 12$. 4. Given that $t_6 + t_{10} = 44$, so $(a + 5d) + (a + 9d) = 44$, which gives $2a + 14d = 44$ or $a + 7d = 22$. 5. Subtracting the first equation from the second: $2d = 10$, so $d = 5$. 6. Substituting $d = 5$ into $a + 5d = 12$, we get $a + 25 = 12$, so $a = -13$. 7. The first three terms are $a = -13$, $a+d = -8$, and $a+2d = -3$. **Answer:** -13, -8, -3 > Common mistake: Errors in solving simultaneous linear equations for $a$ and $d$. ### Question 9 *3 marks · Short answer* Find the smallest value of $n$ such that the sum of the first $n$ natural numbers is greater than 1,000. **Solution** 1. The sum of the first $n$ natural numbers is given by $S_n = \frac{n(n+1)}{2}$. 2. We need to find the smallest value of $n$ such that $\frac{n(n+1)}{2} > 1000$, which means $n(n+1) > 2000$. 3. Testing integer values near $\sqrt{2000} \approx 44.7$, for $n = 44$: $\frac{44 \times 45}{2} = 990$, which is not greater than 1000. 4. For $n = 45$: $\frac{45 \times 46}{2} = 45 \times 23 = 1035$, which is greater than 1000. **Answer:** 45 > Common mistake: Stopping at $n = 44$ without checking if the sum exceeds 1000. ### Question 10 *3 marks · Short answer* Which term of the GP: 2, 8, 32, ... is 131072? Write the explicit formula as well as the recursive formula for the $n^{\text{th}}$ term. **Solution** 1. The given GP is $2, 8, 32, \dots$, where the first term $a = 2$ and the common ratio $r = \frac{8}{2} = 4$. 2. The explicit formula for the $n^{\text{th}}$ term is $t_n = a r^{n-1} = 2 \times 4^{n-1}$. 3. The recursive formula is $t_1 = 2$ and $t_n = 4 t_{n-1}$ for $n \ge 2$. 4. To find which term is 131072, set $t_n = 131072$, so $2 \times 4^{n-1} = 131072$. 5. This simplifies to $4^{n-1} = 65536$. Since $65536 = 4^8$, we have $n - 1 = 8$, so $n = 9$. **Answer:** 9th term, explicit formula: $t_n = 2 \times 4^{n-1}$, recursive formula: $t_1 = 2, t_n = 4t_{n-1}$ for $n \ge 2$. > Common mistake: Equating $4^{n-1}$ directly to the term value instead of dividing by $a$ first. ### Question 11 *3 marks · Short answer* The sum of the first three terms of a GP is $\frac{13}{12}$ and their product is $-1$. Find the common ratio and the terms. **Solution** 1. Let the first three terms of the GP be $\frac{a}{r}$, $a$, and $ar$. 2. The product of the terms is given as $-1$, so $\left(\frac{a}{r}\right)(a)(ar) = -1$, which gives $a^3 = -1$, so $a = -1$. 3. The sum of the terms is $\frac{13}{12}$, so $\frac{a}{r} + a + ar = \frac{13}{12}$. 4. Substitute $a = -1$: $-\frac{1}{r} - 1 - r = \frac{13}{12}$, which gives $-\left(r + \frac{1}{r}\right) = \frac{25}{12}$ or $r + \frac{1}{r} = -\frac{25}{12}$. 5. Multiplying by $12r$, we get $12r^2 + 25r + 12 = 0$. 6. Factorising the quadratic equation: $(4r + 3)(3r + 4) = 0$, so $r = -\frac{3}{4}$ or $r = -\frac{4}{3}$. 7. For $r = -\frac{3}{4}$ and $a = -1$, the terms are $\frac{4}{3}$, $-1$, and $\frac{3}{4}$. (The same set of terms is obtained for $r = -\frac{4}{3}$). **Answer:** Common ratio is $-\frac{3}{4}$ or $-\frac{4}{3}$, and the terms are $\frac{4}{3}$, $-1$, and $\frac{3}{4}$. > Common mistake: Failing to consider both positive and negative roots for the quadratic equation in $r$. ### Question 12 *4 marks · Proof* If the $4^{\text{th}}$, $10^{\text{th}}$ and $16^{\text{th}}$ terms of a GP are $x$, $y$ and $z$ respectively, prove that $x$, $y$, $z$ are in GP. **Solution** 1. Let the first term of the GP be $A$ and the common ratio be $R$. 2. Given that the $4^{\text{th}}$, $10^{\text{th}}$, and $16^{\text{th}}$ terms are $x$, $y$, and $z$ respectively. 3. Therefore, $x = t_4 = A R^3$. 4. Similarly, $y = t_{10} = A R^9$. 5. And $z = t_{16} = A R^{15}$. 6. Now, consider the ratio $\frac{y}{x} = \frac{A R^9}{A R^3} = R^6$. 7. Consider the ratio $\frac{z}{y} = \frac{A R^{15}}{A R^9} = R^6$. 8. Since $\frac{y}{x} = \frac{z}{y} = R^6$, which is a constant, $x, y, z$ form a geometric progression. 9. Hence proved. **Answer:** Hence proved that x, y, z are in GP. > Common mistake: Using wrong indices for the terms of the GP (e.g., writing $AR^4$ for the 4th term). ### Question 13 *3 marks · Short answer* The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP. **Solution** 1. Let the first three terms of the geometric progression be $\frac{a}{r}$, $a$, and $ar$, where $a$ is the first term and $r$ is the common ratio. 2. Given that the sum of the first three terms is 26, we have $\frac{a}{r} + a + ar = 26$, which gives $a(1 + r + r^2) = 26r \quad \text{--- (1)}$ 3. Given that the sum of their squares is 364, we have $\left(\frac{a}{r}\right)^2 + a^2 + (ar)^2 = 364$, which gives $a^2(1 + r^2 + r^4) = 364r^2 \quad \text{--- (2)}$ 4. Squaring equation (1), we get $a^2(1 + r + r^2)^2 = 676r^2$. 5. Dividing this by equation (2) using the algebraic identity $(1 + r + r^2)(1 - r + r^2) = 1 + r^2 + r^4$, we obtain $\frac{(1 + r + r^2)^2}{1 + r^2 + r^4} = \frac{676r^2}{364r^2} = \frac{13}{7}$, which simplifies to $3r^2 - 10r + 3 = 0$. 6. Solving $3r^2 - 10r + 3 = 0$ gives $r = 3$ or $r = \frac{1}{3}$. 7. Substituting $r = 3$ into equation (1) gives $a = 6$, yielding the terms $2, 6, 18$. For $r = \frac{1}{3}$, we get $a = 18$, yielding the terms $18, 6, 2$. **Answer:** The terms of the GP are $2, 6, 18$ or $18, 6, 2$. > Common mistake: Making algebraic errors while squaring and dividing equations for a three-term GP. ### Question 14 *3 marks · Short answer* Suppose $P_1 = 1$, $P_2 = 2$ and for $n > 2$, $P_n = P_1 + P_2 + \dots + P_{n-1} + 1$. Find the values of $P_1$, $P_2$, ..., $P_8$. Can you find a simpler recursive formula for $P_n$? Can you give an explicit formula? **Solution** 1. Given $P_1 = 1$, $P_2 = 2$, and $P_n = P_1 + P_2 + \dots + P_{n-1} + 1$ for $n > 2$. 2. For $n = 3$, $P_3 = P_1 + P_2 + 1 = 1 + 2 + 1 = 4$. 3. For $n = 4$, $P_4 = P_1 + P_2 + P_3 + 1 = 1 + 2 + 4 + 1 = 8$, and similarly $P_5 = 16$, $P_6 = 32$, $P_7 = 64$, and $P_8 = 128$. 4. For the simpler recursive formula, since $P_{n-1} = P_1 + P_2 + \dots + P_{n-2} + 1$, we can write $P_n = P_{n-1} + P_{n-1} = 2P_{n-1}$ for $n \ge 3$ with $P_1 = 1$, $P_2 = 2$. 5. Using the recursive formula $P_n = 2P_{n-1}$, the explicit formula is $P_n = 2^{n-1}$ for $n \ge 1$. **Answer:** Values are $1, 2, 4, 8, 16, 32, 64, 128$; recursive formula is $P_n = 2P_{n-1}$; explicit formula is $P_n = 2^{n-1}$. > Common mistake: Failing to simplify the summation in the recursive definition to a single previous term. ### Question 15 *3 marks · Short answer* Suppose $W_1 = 1$, $W_2 = 2$ and for $n > 2$, $W_n = W_1 + W_2 + \dots + W_{n-2} + 2$. Find the values of $W_1$, $W_2$, ..., $W_8$. Do you recognise this sequence? **Solution** 1. Given $W_1 = 1$, $W_2 = 2$, and $W_n = W_1 + W_2 + \dots + W_{n-2} + 2$ for $n > 2$. 2. For $n = 3$, $W_3 = W_1 + 2 = 1 + 2 = 3$. 3. For $n = 4$, $W_4 = W_1 + W_2 + 2 = 1 + 2 + 2 = 5$. 4. Continuing this for the subsequent terms, we get $W_5 = 8$, $W_6 = 13$, $W_7 = 21$, and $W_8 = 34$. 5. This sequence is the famous Virahānka–Fibonacci sequence where each term is obtained by adding the previous two terms. **Answer:** The values are $1, 2, 3, 5, 8, 13, 21, 34$, and the sequence is the Virahānka–Fibonacci sequence. > Common mistake: Misinterpreting the limits of the summation in the recursive definition. ## Frequently asked questions ### How many exercises and questions are there in Class 9 Maths Chapter 8 according to the new NCERT book? The chapter includes various targeted practice sets along with Exercise Set 8.1 having 6 questions, Exercise Set 8.2 having 7 questions, Exercise Set 8.3 having 7 questions, and End-of-Chapter Exercises featuring 15 questions. You can find complete step-by-step solutions and the free PDF for all these questions right here on this SwaVid page. ### What main topics and concepts are covered in the exercises of this chapter? The questions cover sequences, triangular numbers, explicit rules like $u_n = 2n - 1$, recursive sequences, Arithmetic Progressions, Geometric Progressions, and sums of consecutive natural numbers. These concepts help students learn how to predict what comes next in various patterns. ### Which question types are considered the most challenging in this chapter and how should we approach them? The proof-based questions and long answer problems found in the End-of-Chapter Exercises and Exercise Set 8.3 are generally the hardest. To score full marks, students should carefully analyze the recursive or explicit formulas, break down word problems step by step, and apply AP or GP properties methodically. ### How can I write my answers to get full marks in Class 9 exams for sequence and progression problems? To secure full marks, always state the given rule or formula clearly, show every substitution step like $t_n = a + (n - 1)d$, and write proper units where applicable. Following the detailed solutions provided in SwaVid's free PDF on this page will help you understand the correct presentation style. ### Is the free PDF for Class 9 Maths Chapter 8 solutions available for the 2026-27 session? Yes, complete solutions aligned with the new NCERT book for the 2026-27 session are available on this SwaVid page. You can easily access the free PDF to practice concepts ranging from basic sequences to advanced geometric progressions. ## Related pages - [Exercise 8.1 solutions](https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-1) - [Exercise 8.2 solutions](https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-2) - [Exercise 8.3 solutions](https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-3) - [Class 9 Maths chapters](https://www.swavid.com/maths/class/9) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student