Ganita Manjari Class 9 Maths Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions — NCERT Solutions

Explanation: The counts 3, 6, 12, 24 double at every stage, so the pattern is a geometric progression with first term 3 and common ratio 2, giving Stage n = 3 x 2^(n-1). So Stage 5 = 24 x 2 = 48 and Stage 6 = 96. Stage 10 = 3 x 2⁹ = 1536, Stage 11 = 3072, Stage 12 = 6144, and Stage 20 = 3 x 2¹⁹ = 3 x 524288 = 1572864. In Fig. 8.3 the number of squares grows by adding the same number (4) at each stage, which is slow linear growth, whereas here each stage multiplies by 2, so the count grows far faster (exponential growth).

Explanation: Fig. 8.1 shows the first five triangular numbers 1, 3, 6, 10, 15, where the nth pattern is a triangle of dots with rows containing 1, 2, 3, ..., n dots. Each new term adds one more row with one extra dot, so the differences are 2, 3, 4, 5, and the next differences are 6 and 7. Hence the 6th pattern is a triangle with rows of 1 to 6 dots, giving 15 + 6 = 21 dots, and the 7th pattern has rows of 1 to 7 dots, giving 21 + 7 = 28 dots.

Explanation: The sequence 1, 5, 9, 13 increases by 4 at every stage, so it is an arithmetic progression with first term 1 and common difference 4. The rule is Stage n = 1 + 4(n - 1) = 4n - 3. Substituting: Stage 5 = 17, Stage 6 = 21, Stage 10 = 37, Stage 11 = 41, Stage 12 = 45, and Stage 20 = 4 x 20 - 3 = 77. The rule makes sense because each new stage adds one square to each of the four arms of the figure.

Explanation: The list shows four sequences. Natural numbers 1, 2, 3, 4, 5, 6, ... grow by adding 1 each time, so the next terms are 7, 8, 9. Odd numbers 1, 3, 5, 7, 9, 11, ... grow by adding 2, giving 13, 15, 17. Triangular numbers 1, 3, 6, 10, 15, 21, ... have differences that increase by 1 each time (2, 3, 4, 5, 6), so the next differences are 7, 8, 9, giving 28, 36, 45; the nth term is n(n+1)/2. Square numbers 1, 4, 9, 16, 25, 36, ... are the perfect squares n² (their differences are the odd numbers 3, 5, 7, ...), so the next terms are 7² = 49, 8² = 64, 9² = 81.

Explanation: In Fig. 8.2 each square number is drawn as an n x n array of dots, and the array is split into L-shaped borders. The innermost corner has 1 dot, and each successive L-shaped border contains the next odd number of dots: 3, 5, 7, and so on. Adding an L-shaped border of the next odd number of dots to an n x n square turns it into an (n+1) x (n+1) square. So 1 = 1², 1 + 3 = 4 = 2², 1 + 3 + 5 = 9 = 3², 1 + 3 + 5 + 7 = 16 = 4², showing that the sum of the first n odd numbers is exactly n².

Explanation: Using u_n = 2u_(n-1) + 3 with u_1 = 1: u_2 = 2(1) + 3 = 5, u_3 = 2(5) + 3 = 13, u_4 = 2(13) + 3 = 29. Continuing the same rule, u_5 = 2(29) + 3 = 61, u_6 = 2(61) + 3 = 125, u_7 = 2(125) + 3 = 253. The terms are strictly increasing and the sequence jumps from 125 directly to 253, so no term can equal 133. Hence 133 is not a term of the sequence.

Explanation: Substitute each value of n into t_n = n(n+1)/2. For n = 10: t_10 = (10 x 11)/2 = 110/2 = 55. For n = 17: t_17 = (17 x 18)/2 = 306/2 = 153. For n = 80: t_80 = (80 x 81)/2 = 6480/2 = 3240.

Explanation: A sequence is a geometric progression when the ratio of every term to the previous term is the same constant. Here 2/1 = 4/2 = 8/4 = 16/8 = 2, so the ratio is constant throughout. Hence the sequence is a GP with first term 1 and common ratio 2; each term is obtained by multiplying the previous term by 2.

Explanation: An arithmetic progression must have the same difference between every pair of consecutive terms. Natural numbers (difference 1), odd numbers (difference 2), and the pattern 1, 5, 9, 13, ... from Fig. 8.3 (difference 4) all satisfy this, so they are APs. Triangular numbers 1, 3, 6, 10, 15, ... are not, because their differences 2, 3, 4, 5, ... keep increasing; square numbers 1, 4, 9, 16, ... are not, because their differences 3, 5, 7, ... also increase. Sequences such as 3, 6, 12, 24, ... and 1, 2, 4, 8, 16, ... and 1, 3, 9, 27, 81, ... are not APs either, since their differences keep doubling or tripling; they instead have a constant ratio between terms, which makes them geometric progressions.

Explanation: Substitute n = 1, 2, 3, 4, 5, 6 into s_n = 5n - 2: s_1 = 5(1) - 2 = 3, s_2 = 8, s_3 = 13, s_4 = 18, s_5 = 23, s_6 = 28, which form an AP with common difference 5. The advantage of an explicit formula is that any term can be found directly: s_1000 = 5 x 1000 - 2 = 5000 - 2 = 4998.

Explanation: A finite sequence is an ordered list that has a definite first term and a definite last term. Everyday examples include the days of the week from Monday to Sunday (7 terms), the months of the year (12 terms), the page numbers of a book (1 up to the last page), the roll numbers of students in a class, and the number of days in each of the 12 months of a year. Each of these lists is written in a fixed order and stops after finitely many terms.

Explanation: Apply the rule s_n = s_(n-1)(s_(n-1) - 1) starting from s_1 = 3. s_2 = 3 x (3 - 1) = 3 x 2 = 6. s_3 = 6 x (6 - 1) = 6 x 5 = 30. s_4 = 30 x (30 - 1) = 30 x 29 = 870. So the first four terms are 3, 6, 30, 870.

Explanation: Yes, the same pairing method works. Pair the numbers from the two ends of the series: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on up to 50 + 51 = 101. The 100 numbers form exactly 50 such pairs, each with sum 101, so the total is 50 x 101 = 5050. In general this method gives 1 + 2 + ... + n = n(n+1)/2.

Explanation: Check the ratio of each term to the term before it: 3/1 = 3, 9/3 = 3, 27/9 = 3, 81/27 = 3. Since the ratio is the same constant throughout, the sequence is a geometric progression with first term 1 and common ratio 3; each term is obtained by multiplying the previous term by 3.