Call : 94779 94779
GANITA MANJARI | Chapter 8 - Sequences and Progressions
CHAPTER 8

Predicting What Comes Next: Exploring Sequences and Progressions

GANITA MANJARI | Class 9 Maths | Part I

Think and Reflect

Page 174
Can you describe the pattern in each of the above sequences? Can you predict the next few numbers in these sequences?
Answer
Natural numbers: add 1 each time → next: 7,8,9. Odd numbers: add 2 → next: 13,15,17. Triangular numbers: add increasing numbers (2,3,4,5,6,...) → next: 28,36,45. Square numbers: add consecutive odd numbers → next: 49,64,81.
Page 176
Can you think of any other kinds of sequences? List out five different types of sequences and discuss their properties with your friends.
Answer
Examples: Arithmetic progression, geometric progression, Fibonacci sequence, alternating sequences, sequences of powers.
Page 177
Why is it useful to have an explicit formula for the nth term of a sequence?
Answer
It allows direct calculation of any term without knowing previous terms.
Page 180
Can you predict the number of squares in Stages 5 and 6 of the sequence? In Stages 10, 11 and 12? In Stage 20? At any stage?
Answer
Pattern: t_n = 4n - 3. Stage 5: 17, Stage 6: 21, Stage 10: 37, Stage 11: 41, Stage 12: 45, Stage 20: 77.
Page 185
Can you use this formula to find S20, S50 or S1000? Also, this formula can be used to find the sum of consecutive numbers such as 25 + 26 + 27 + ... + 58. Let us revisit the sequence tn of triangular numbers 1, 3, 6, 10, 15, ... shown in Fig. 8.1. Note that the nth term of this sequence is the sum of the first n natural numbers. Thus tn = n(n+1)/2. Can you use this to find the 10th, 17th and 80th triangular numbers?
Answer
S20 = 210, S50 = 1275, S1000 = 500500. Triangular numbers: t10 = 55, t17 = 153, t80 = 3240.

EXERCISE SET 8.1 (Pages 179-180)

1.
Find the first five terms of the sequence in which the nth term is given by (i) t_n = 3n - 4, (ii) t_n = 2 - 5n, and (iii) t_n = n² - 2n + 3 for n ≥ 1.
Solution
(i) n=1→-1, 2→2, 3→5, 4→8, 5→11
(ii) n=1→-3, 2→-8, 3→-13, 4→-18, 5→-23
(iii) n=1→2, 2→3, 3→6, 4→11, 5→18
2.
Find the 10th and 15th terms of the sequence t_n = 5n - 3 for n ≥ 1.
Solution
t10 = 5×10 - 3 = 47, t15 = 5×15 - 3 = 72
3.
Determine whether 97 and 172 are terms of the sequence t_n = 5n - 3 for n ≥ 1.
Solution
5n - 3 = 97 ⇒ 5n = 100 ⇒ n = 20 (yes). 5n - 3 = 172 ⇒ 5n = 175 ⇒ n = 35 (yes).
4.
Which term of the sequence t_n = 5n - 3 for n ≥ 1 is 607?
Solution
5n - 3 = 607 ⇒ 5n = 610 ⇒ n = 122. So 122nd term.
5.
A sequence is given by the recursive rule t₁ = -5, t_{n+1} = t_n + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
Solution
Terms: -5, -2, 1, 4, 7. For 52: t_n = -5 + (n-1)×3 = 52 ⇒ 3(n-1) = 57 ⇒ n-1 = 19 ⇒ n = 20. Yes, 20th term.
6.
Let T₁ = 1, T₂ = 2, T₃ = 4 and T_n = T_{n-1} + T_{n-2} + T_{n-3} for n ≥ 4. Find T₄, T₅, T₆, T₇ and T₈.
Solution
T₄ = 4+2+1 = 7, T₅ = 7+4+2 = 13, T₆ = 13+7+4 = 24, T₇ = 24+13+7 = 44, T₈ = 44+24+13 = 81.

EXERCISE SET 8.2 (Pages 185-186)

1.
Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ...
Solution
a = 3, d = 5. t10 = 3 + 9×5 = 48, t26 = 3 + 25×5 = 128.
2.
Which term of the AP: 21, 18, 15, ... is -81? Also, is 0 a term of this AP? Give reasons for your answer.
Solution
a = 21, d = -3. t_n = 21 + (n-1)(-3) = -81 ⇒ 21 - 3n + 3 = -81 ⇒ 24 - 3n = -81 ⇒ -3n = -105 ⇒ n = 35. 0: 24 - 3n = 0 ⇒ 3n = 24 ⇒ n = 8. Yes, 0 is the 8th term.
3.
Find the nth term of the AP: 11, 8, 5, 2, ... Write the recursive rule for this AP.
Solution
a = 11, d = -3. t_n = 11 + (n-1)(-3) = 14 - 3n. Recursive: t₁ = 11, t_n = t_{n-1} - 3 for n ≥ 2.
4.
An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Solution
a + 2d = 12, a + 49d = 106. Subtract: 47d = 94 ⇒ d = 2, then a = 8. t29 = 8 + 28×2 = 64.
5.
How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Solution
First: 12, last: 99. Number of terms = ((99-12)/3) + 1 = (87/3)+1 = 29+1 = 30. Sum = (30/2)(12+99) = 15 × 111 = 1665.
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?
Solution
5,00,000 + (n-1)×20,000 = 7,00,000 ⇒ (n-1)×20,000 = 2,00,000 ⇒ n-1 = 10 ⇒ n = 11 years.
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?
Solution
Sum = 1+2+...+25 = 25×26/2 = 325 marbles.

EXERCISE SET 8.3 (Pages 193-194)

1.
Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Solution
t8 = a × 2⁷ = 192 ⇒ a × 128 = 192 ⇒ a = 192/128 = 3/2. t12 = (3/2) × 2¹¹ = (3/2) × 2048 = 3 × 1024 = 3072.
2.
Find the 10th and nth terms of the GP: 5, 25, 125, ...
Solution
a = 5, r = 5. t10 = 5 × 5⁹ = 5¹⁰ = 9765625. t_n = 5 × 5^{n-1} = 5^n.
*3.
A sequence is given by the recursive rule t₁ = 2, t_{n+1} = 3t_n - 2 for n ≥ 1. Which term of the sequence is 730?
Solution
t₁=2, t₂=4, t₃=10, t₄=28, t₅=82, t₆=244, t₇=730. So 730 is the 7th term.
4.
Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the nth term.
Solution
a=2, r=3. t_n = 2 × 3^{n-1} = 4374 ⇒ 3^{n-1} = 2187 = 3⁷ ⇒ n-1=7 ⇒ n=8. Explicit: t_n = 2 × 3^{n-1}. Recursive: t₁=2, t_n = 3t_{n-1} for n≥2.
5.
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 5th bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Solution
(i) After 5th bounce: 80 × (0.6)⁵ = 80 × 0.07776 = 6.2208 m.
(ii) Total = 80 + 2×80(0.6 + 0.6² + 0.6³ + 0.6⁴ + 0.6⁵) = 80 + 160×0.6(1-0.6⁵)/(1-0.6) = 80 + 96×(1-0.07776)/0.4 = 80 + 240×0.92224 = 80 + 221.3376 = 301.34 m approx.
6.
Which term of the sequence √2, 2, 2√2, 4, ... is 128?
Solution
a = √2, r = √2. t_n = √2 × (√2)^{n-1} = (√2)^n = 2^{n/2} = 128 = 2⁷ ⇒ n/2 = 7 ⇒ n = 14. So 14th term.
7.
Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. 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 nth 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 nth stage. What happens to this area as n, the number of stages, goes on increasing?
Solution
(i) Stage 0: 1, Stage 1: 8, Stage 2: 64, Stage 3: 512 red squares.
(ii) Stage 4: 4096, Stage 5: 32768.
(iii) Explicit: R_n = 8^n. Recursive: R₀ = 1, R_n = 8 × R_{n-1} for n ≥ 1.
(iv) Area: Stage 0: 1, Stage 1: 8/9, Stage 2: 64/81, Stage 3: 512/729. Stage 4: 4096/6561, Stage 5: 32768/59049. Explicit: A_n = (8/9)^n. As n increases, area approaches 0.

END-OF-CHAPTER EXERCISES (Pages 194-195)

1.
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution
a + 10d = 38, a + 15d = 73. Subtract: 5d = 35 ⇒ d = 7, then a = 38 - 70 = -32. t31 = -32 + 30×7 = -32 + 210 = 178.
2.
Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
Solution
a + 2d = 16. t7 - t5 = (a+6d) - (a+4d) = 2d = 12 ⇒ d = 6. Then a = 16 - 12 = 4. AP: 4, 10, 16, 22, ...
*3.
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
Smallest: 105, largest: 994. Number of terms = ((994-105)/7) + 1 = (889/7)+1 = 127+1 = 128.
*4.
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
Smallest: 12, largest: 248. Number of terms = ((248-12)/4) + 1 = (236/4)+1 = 59+1 = 60.
*5.
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
a + ar = -4 ⇒ a(1+r) = -4. t5 = 4×t3 ⇒ ar⁴ = 4×ar² ⇒ r² = 4 ⇒ r = 2 or r = -2. If r=2: a(3) = -4 ⇒ a = -4/3. GP: -4/3, -8/3, -16/3, ... If r=-2: a(1-2) = -4 ⇒ a(-1) = -4 ⇒ a = 4. GP: 4, -8, 16, -32, ...
*6.
Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Solution
100 = 9+10+11+12+13+14+15+16 (8 numbers), = 18+19+20+21+22 (5 numbers), = 100 itself (1 number).
*7.
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 2nd hour, 4th hour and nth hour?
Solution
At end of 2nd hour: 30×2² = 120. At end of 4th hour: 30×2⁴ = 480. At end of nth hour: 30×2ⁿ.
*8.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution
(a+3d)+(a+7d)=2a+10d=24 ⇒ a+5d=12. (a+5d)+(a+9d)=2a+14d=44 ⇒ a+7d=22. Subtract: 2d=10 ⇒ d=5, then a=12-25=-13. AP: -13, -8, -3, ...
*9.
Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
Solution
n(n+1)/2 > 1000 ⇒ n(n+1) > 2000. n=44 → 44×45=1980 <2000, n=45 → 45×46=2070 >2000. So n=45.
*10.
Which term of the GP: 2, 8, 32, ... is 131072? Write the explicit formula as well as the recursive formula for the nth term.
Solution
a=2, r=4. t_n = 2 × 4^{n-1} = 131072 = 2¹⁷ ⇒ 4^{n-1} = 2¹⁶ = 4⁸ ⇒ n-1=8 ⇒ n=9. Explicit: t_n = 2 × 4^{n-1}. Recursive: t₁=2, t_n = 4t_{n-1} for n≥2.
*11.
The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the terms.
Solution
Let terms be a/r, a, ar. Sum = a(1/r + 1 + r) = 13/12. Product = a³ = -1 ⇒ a = -1. Then (-1)(1/r + 1 + r) = 13/12 ⇒ 1/r + 1 + r = -13/12 ⇒ 1/r + r = -25/12. Multiply by 12r: 12 + 12r² = -25r ⇒ 12r² + 25r + 12 = 0 ⇒ (3r+4)(4r+3)=0 ⇒ r = -4/3 or r = -3/4. Terms: if r=-4/3: 3/4, -1, 4/3. If r=-3/4: 4/3, -1, 3/4.
*12.
If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
Solution
t4 = ar³ = x, t10 = ar⁹ = y, t16 = ar¹⁵ = z. Then y² = a²r¹⁸ = (ar³)(ar¹⁵) = xz. Hence x, y, z are in GP.
*13.
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
Let terms be a/r, a, ar. Sum = a(1/r + 1 + r) = 26. Sum of squares = a²(1/r² + 1 + r²) = 364. Let S = a(1/r + r) and P = a. Then (1/r + r)² = 1/r² + r² + 2. Solving gives a=6, r=2 or r=1/2. Terms: 3,6,12 or 12,6,3.
*14.
Suppose P₁ = 1, P₂ = 2 and for n > 2, P_n = P₁ + P₂ + ... + P_{n-1} + 1. Find the values of P₁, P₂, ..., P₈. Can you find a simpler recursive formula for P_n? Can you give an explicit formula?
Solution
P₁=1, P₂=2, P₃=1+2+1=4, P₄=1+2+4+1=8, P₅=1+2+4+8+1=16, P₆=32, P₇=64, P₈=128. Recursive: P_n = 2P_{n-1} for n≥3. Explicit: P₁=1, P_n = 2^{n-2} for n≥2.
*15.
Suppose W₁ = 1, W₂ = 2 and for n > 2, W_n = W₁ + W₂ + ... + W_{n-2} + 2. Find the values of W₁, W₂, ..., W₈. Do you recognise this sequence?
Solution
W₁=1, W₂=2, W₃=1+2=3? Wait: W₃ = W₁ + 2 = 1+2=3, W₄ = W₁+W₂+2 = 1+2+2=5, W₅ = W₁+W₂+W₃+2 = 1+2+3+2=8, W₆ = 1+2+3+5+2=13, W₇ = 1+2+3+5+8+2=21, W₈ = 1+2+3+5+8+13+2=34. This is the Fibonacci sequence (starting 1,2,3,5,8,13,21,34,...).

Chapter Summary (Page 196)

  • A sequence is an ordered list of numbers. Each number is called a term.
  • An explicit formula uses the term's position number n to calculate its value.
  • A recursive formula gives the value of a term using previous terms.
  • Triangular numbers: t_n = n(n+1)/2.
  • Arithmetic Progression (AP): t_n = a + (n-1)d, where d is the common difference.
  • Geometric Progression (GP): t_n = ar^{n-1}, where r is the common ratio.
  • Sum of first n natural numbers: S_n = n(n+1)/2.
  • Fractals like Sierpiński triangle and square carpet lead to geometric progressions.

Also Get

Class 9- NCERT- Science Solⁿ.

Class 9-NCERT- English Solⁿ

Class 9-NCERT- Maths
Solⁿ

Class 9-NCERT- Social Science Solⁿ

Class 9 CBSE - SYLLABUS

Study materials

Send Us A Message






    Latest posts

    Facebook Twitter Youtube Icon-instagram-1

    Thank You For Registering !

    Our Team Will Reach You Soon

    Enter Your Contact Details To Download the PDF

      This form uses Akismet to reduce spam. Learn how your data is processed.

      REGISTER NOW

      👉 (Limited Seats Available) 👈