Sequences and progressions: Questions and Answers

75 sequences and progressions questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• Can you describe the pattern in each of the above sequences? Can you predict the next few numbers in these sequences?Answer: (i) Add 1 each time: 7, 8, 9, … (ii) Add 2 each time: 13, 15, 17, … (iii) Add 2, then 3, then 4, …: 28, 36, 45, … (iv) Add 3, 5, 7, … (the squares 1², 2², 3², …): 49, 64, 81, …
• The sequence 6, 12, 24, 48, 96 is a finite sequence of five terms. Can you think of other finite sequences that you see in your daily life?Answer: Yes. Examples: the dates of a month 1, 2, 3, …, 30; the roll numbers of a class 1, 2, …, 40; the floor numbers of a 12-storey building 0, 1, 2, …, 11; the page numbers of a book; the scores after each over of a cricket innings.
• Each term of the triangular number sequence is the sum of the natural numbers up to that term. For example, 15, the fifth triangular number, is equal to 1 + 2 + 3 + 4 + 5. This is represented by the diagram in the figure, where each triangular number is represented by a triangular array of dots. Can you draw the patterns for the next two terms of the sequence?Answer: The 6th triangular number is 21 (a triangle with 6 rows: 1 + 2 + 3 + 4 + 5 + 6) and the 7th is 28 (7 rows: 1 + 2 + … + 7).
• 1 = 1, 4 = 1 + 3, 9 = 1 + 3 + 5, 16 = 1 + 3 + 5 + 7, and so on. Each term in the square number sequence is the sum of the odd numbers up to that term. This interesting relationship between the odd numbers and square numbers can be represented by the diagram in the figure. Can you explain the relationship?Answer: To grow a square of dots from side k − 1 to side k, you add an L-shaped band of k + k − 1 = 2k − 1 dots, which is the kth odd number. So 1 + 3 + 5 + … + (2n − 1) = n × n = n².
• Consider the sequence 1, 4, 7, 10, 13, … 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.)Answer: Next four terms: 16, 19, 22, 25. Running-sum sequence (first 10 terms): 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.
• Can you write t₅, t₆, t₇ and t₈ for the sequence of triangular numbers?Answer: t₅ = 15, t₆ = 21, t₇ = 28, t₈ = 36.
• Can you think of any other kinds of sequences? List out five different types of sequences and discuss their properties with your friends.Answer: For example: (1) increasing by a fixed step, 5, 10, 15, …; (2) decreasing, 100, 90, 80, …; (3) multiplying each time, 2, 4, 8, 16, …; (4) alternating signs, 1, −1, 1, −1, …; (5) repeating, 1, 2, 3, 1, 2, 3, …. A sixth kind: each term made from the two before it, 1, 1, 2, 3, 5, 8, ….
• Consider the expression u_n = 2n − 1. This states that the nth term of the sequence is given by the rule 2n − 1.Answer: u₁ = 1, u₂ = 3, u₃ = 5, … so u_n = 2n − 1 is the explicit rule for the sequence of odd numbers 1, 3, 5, 7, ….
• Why is it useful to have an explicit formula for the nth term of a sequence?Answer: Because it gives any term directly (the 20th, the 300th, …) just by putting in n, without working out all the earlier terms. It also lets us check whether a number is a term and find its position (e.g. 2n − 1 = 137 gives n = 69).
• Using the explicit rule u_n = 2n − 1, find the 53rd term, the 108th term, and the 1170th term of the odd number sequence.Answer: u₅₃ = 105, u₁₀₈ = 215, u₁₁₇₀ = 2339.
• Consider the sequence that is generated by the explicit formula s_n = 5n − 2.Answer: First 6 terms: 3, 8, 13, 18, 23, 28. 100th term = 498; 1000th term = 4998. 308 is the 62nd term. 471 is not a term (n = 94.6); 473 is the 95th term. n must be a natural number because it counts a position.
• Can you find the rule describing the nth term of the sequence of square numbers?Answer: t_n = n² (that is, n × n).
• Here is the sequence of the first ten prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Do you see any pattern in this sequence? Can you think of a rule that can predict the next few prime numbers?Answer: There is no regular pattern: the gaps are 1, 2, 2, 4, 2, 4, 2, 4, 6. Some facts hold (after 2 every prime is odd; after 3 every prime is 1 more or 1 less than a multiple of 6), but no simple rule gives the next prime. We find them by testing: the next primes are 31, 37, 41, 43, 47.
• Consider the expression t_n = 3n − 7.Answer: (i) −4, −1, 2, 29, 47, 143 (ii) 332 is the 113th term (iii) Yes: 3n − 7 = 557 gives n = 188, a natural number, so 557 is the 188th term.
• Consider the sequence 1, 4, 7, 10, 13, … . The nth term is t_n = 3n − 2 (verify this for yourself).Answer: Putting n = 1, 2, 3, 4, 5 in 3n − 2 gives 1, 4, 7, 10, 13 ✓. Also 3n − 2 goes up by exactly 3 when n goes up by 1, just as the sequence does.
• Find the first four terms of the sequence given by the recursive rule u₁ = 1, u_n = 2u_n−1 + 3 for n ≥ 2. Is 133 a term of this sequence?Answer: First four terms: 1, 5, 13, 29. Continuing: 61, 125, 253, … 133 lies between 125 and 253, so 133 is not a term.
• Find the first four terms of the sequence given by the recursive rule s₁ = 3, s_n = s_n−1 (s_n−1 − 1) for n ≥ 2.Answer: 3, 6, 30, 870
• A recursive rule or formula does not only have to involve the previous term — it could involve the previous two or more terms. The most famous example of such a sequence is V₁, V₂, V₃, … where V₁ = 1, V₂ = 2, and V_n = V_n−1 + V_n−2 for n ≥ 3. So we get the sequence 1, 2, 3, 5, 8, 13, 21, 34, … where each term is obtained by adding the previous two. Can you write the next two terms of this sequence?Answer: 55 and 89 (21 + 34 = 55, 34 + 55 = 89). This is the Virahānka–Fibonacci sequence.
• Find the first five terms of the sequence in which the nth term is given byAnswer: (i) −1, 2, 5, 8, 11 (ii) −3, −8, −13, −18, −23 (iii) 2, 3, 6, 11, 18
• Find the 10th and 15th terms of the sequence t_n = 5n − 3 for n ≥ 1.Answer: t₁₀ = 47, t₁₅ = 72.
• Determine whether 97 and 172 are terms of the sequence t_n = 5n − 3 for n ≥ 1.Answer: Both are terms: 97 is the 20th term and 172 is the 35th term.
• Which term of the sequence t_n = 5n − 3 for n ≥ 1 is 607?Answer: 607 is the 122nd term.
• 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?Answer: First five terms: −5, −2, 1, 4, 7. Explicit form t_n = 3n − 8; 3n − 8 = 52 gives n = 20, so 52 is the 20th term.
• 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.Answer: T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81.
• Let us look at the growing pattern of squares given in the figure. The first four stages of the pattern are shown. If we count the number of tiny squares at each stage, we get sequence 1, 5, 9, 13. 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: Stage 5: 17, Stage 6: 21; Stages 10, 11, 12: 37, 41, 45; Stage 20: 77; Stage n: 4n − 3.
• Consider all the sequences we have discussed so far. Which ones are arithmetic progressions and which ones are not? Can you justify your claim?Answer: APs (constant difference d): natural numbers (d = 1), odd numbers (d = 2), 1, 4, 7, 10, … (d = 3), −7, −3, 1, 5, … (d = 4), 5n − 2 (d = 5), 3n − 7 (d = 3), 3n − 4 (d = 3), 2 − 5n (d = −5), 5n − 3 (d = 5), −5, −2, 1, 4, … (d = 3), 1, 5, 9, 13, … (d = 4). Not APs (differences change): triangular numbers, square numbers, 6, 12, 24, 48, 96, 1, 12, 13, …, the primes, 1, 5, 13, 29, …, 3, 6, 30, 870, the Virahānka–Fibonacci sequence, n² − 2n + 3, the running sums 1, 5, 12, 22, … and 1, 2, 4, 7, 13, ….
• Verify that the following sequences are arithmetic progressions and write their nth terms. What do you observe when you plot the ordered pairs emerging from them?Answer: (i) AP with d = 3; t_n = 3n − 1 (ii) AP with d = 4; t_n = 4n − 9. In both cases the points (n, t_n) lie on a straight line.
• Using the formula t_n = a + (n − 1) × d, find the nth term of the following arithmetic progressions.Answer: (i) t_n = 2n − 32 = 4n − 32 (ii) t_n = 2n − 0.5
• Consider once again the AP: 1, 5, 9, 13, 17, … Since every term is 4 more than the previous term, another way of writing this sequence is t₁ = 1, t_n = t_n−1 + 4 for n ≥ 2. Verify for yourself that this recursive rule leads to the terms of the sequence.Answer: t₁ = 1, t₂ = 1 + 4 = 5, t₃ = 5 + 4 = 9, t₄ = 13, t₅ = 17 ✓. The recursive rule gives exactly 1, 5, 9, 13, 17, ….
• Find recursive rules for the APs in the previous exercises.Answer: 2, 5, 8, …: t₁ = 2, t_n = t_n−1 + 3. −5, −1, 3, …: t₁ = −5, t_n = t_n−1 + 4. 12, 52, …: t₁ = 12, t_n = t_n−1 + 2. 1.5, 3.5, …: t₁ = 1.5, t_n = t_n−1 + 2. (All for n ≥ 2.)
• A person books a taxi to travel in the city. The taxi company charges a fixed booking fee of ₹200 plus ₹40 per kilometre travelled. Let us write the sequence representing the total fare after travelling 1 km, 2 km, 3 km, and so on. If the person travels 10 km, what will be the total fare?Answer: Fares: ₹240, ₹280, ₹320, … an AP with first term 240 and common difference 40; fare for n km = 200 + 40n. For 10 km the fare is ₹600.
• Can you find the sum of the first ten natural numbers without actually adding all of them?Answer: Yes: write the sum forwards and backwards; each of the 10 pairs adds to 11, so 2S = 10 × 11 = 110 and S = 55.
• Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?Answer: Yes. Writing the sum forwards and backwards gives 100 pairs each adding to 101, so 2S = 100 × 101 = 10100 and S = 5050.
• If we use the notation S_n to represent the sum of the first n natural numbers, then S_n = n(n + 1)2. Can you use this formula to find S₂₀, S₅₀ or S₁₀₀₀?Answer: S₂₀ = 210, S₅₀ = 1275, S₁₀₀₀ = 500500.
• Let us revisit the sequence t_n of triangular numbers 1, 3, 6, 10, 15, … shown in the figure. Note that the nth term of this sequence is the sum of the first n natural numbers. Thus t_n = n(n + 1)2. Can you use this to find the 10th, 17th and 80th triangular numbers?Answer: t₁₀ = 55, t₁₇ = 153, t₈₀ = 3240.
• Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….Answer: t₁₀ = 48, t₂₆ = 128.
• Which term of the AP : 21, 18, 15, … is −81? Also, is 0 a term of this AP? Give reasons for your answer.Answer: −81 is the 35th term. Yes, 0 is a term: 21 − 3(n − 1) = 0 gives n = 8, a natural number, so 0 is the 8th term.
• Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.Answer: t_n = 14 − 3n. Recursive rule: t₁ = 11, t_n = t_n−1 − 3 for n ≥ 2.
• An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th 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’.)Answer: a = 8, d = 2, so the 29th term = 8 + 28 × 2 = 64.
• How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?Answer: 30 two-digit numbers (12, 15, …, 99) are divisible by 3. Their sum is 1665.
• 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?Answer: After 10 years (10 increments of ₹20,000). His salary in the 11th year of work is ₹7,00,000.
• 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?Answer: 1 + 2 + … + 25 = 25 × 262 = 325 marbles.
• Consider the growing pattern of squares shown in the figure. The first four stages of the pattern are shown. If we count the total number of green squares in the four stages of the pattern, we get the sequence 3, 6, 12, 24. Can you predict the number of squares in Stages 5 and 6 of the pattern? In Stages 10, 11 and 12? In Stage 20? At any stage? How is this different from the growing pattern in the figure?Answer: Stages 5, 6: 48, 96; Stages 10, 11, 12: 1536, 3072, 6144; Stage 20: 1572864; Stage n: 3 × 2ⁿ⁻¹. Here the number doubles each stage (multiply by 2), while the earlier pattern 1, 5, 9, 13 adds 4 each stage, so this one grows far faster.
• Is 1, 2, 4, 8, 16, … a geometric progression? If so, what is the common ratio?Answer: Yes, it is a GP: every term divided by the term before it gives 2. Common ratio r = 2.
• Is 1, 3, 9, 27, 81, … a geometric progression? If so, what is the common ratio?Answer: Yes, it is a GP: every term divided by the term before it gives 3. Common ratio r = 3.
• Is 1, −1, 1, −1, 1, … a geometric progression? If so, what is the common ratio?Answer: Yes, it is a GP: every term divided by the term before it gives −1. Common ratio r = −1.
• Check whether the sequence 5, 154, 4516, 13564, … is a geometric progression and find its nth term.Answer: Every ratio is 34, so it is a GP with a = 5, r = 34. nth term = 5 × (34)ⁿ⁻¹.
• Check whether the following sequences are geometric progressions and find their nth terms.Answer: All three are GPs. (i) r = 5, t_n = 2 × 5ⁿ⁻¹ (ii) r = 23, t_n = 4 × (23)ⁿ⁻¹ (iii) r = −12, t_n = 3 × (−12)ⁿ⁻¹
• Can you find a recursive rule for the formula t_n = 3 × 10ⁿ⁻¹ that generates the geometric progression 3, 30, 300, 3000, … ?Answer: t₁ = 3, t_n = 10t_n−1 for n ≥ 2.
• Observe the Sierpiński triangle and try to answer the following questionsAnswer: (a) 1, 3, 9, 27 (b) 81 and 243 (c) 3ⁿ (d) 34, 916, 2764; 81256, 2431024; area = (34)ⁿ, which gets closer and closer to 0 as n increases.
• The number of black triangles at the nth stage of the Sierpiński triangle is given by 3ⁿ. The number of black triangles increases very quickly as the stage numbers increase. Can you explain why?Answer: Because the number is multiplied by 3 at every stage (every black triangle becomes 3), not increased by a fixed amount. Repeated multiplication grows very fast: 1, 3, 9, 27, 81, 243, … and by Stage 10 there are 3¹⁰ = 59049 triangles.
• If the black region at Stage 0 is 1 square unit, then the black region at Stage 1 is 34 square units. This process is repeated on Stage 1 to arrive at Stage 2. Hence, the black region at Stage 2 will be 34 of the black region of Stage 1, which is equal to 34 × 34 = (34)². Can you explain why the area of the black region at Stage n will be (34)ⁿ?Answer: Each stage keeps 3 of the 4 equal pieces of every black triangle, so the black area is multiplied by 34 at every stage. Starting from 1 and multiplying by 34 n times gives (34)ⁿ.
• A ball is dropped from a height of 24 feet above the ground. Each time the ball bounces up to (34)^th of its previous height.Answer: (a) 18, 13.5, 10.125, 7.59375, 5.6953125 (≈ 5.695) feet: a GP with a = 18, r = 34. (b) 16 of 24 = 4 feet; after the 6th bounce the height is about 4.27 ft, after the 7th about 3.20 ft, so 7 bounces.
• Find the 12th term of a GP with common ratio 2, whose 8th term is 192.Answer: 12th term = 192 × 2⁴ = 3072.
• Find the 10th and nth terms of the GP: 5, 25, 125, … .Answer: t_n = 5 × 5ⁿ⁻¹ = 5ⁿ; t₁₀ = 5¹⁰ = 9765625.
• 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?Answer: Terms: 2, 4, 10, 28, 82, 244, 730, … so 730 is the 7th term. (In fact t_n = 3ⁿ⁻¹ + 1, and 730 = 3⁶ + 1.)
• Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.Answer: 4374 is the 8th term. Explicit: t_n = 2 × 3ⁿ⁻¹. Recursive: t₁ = 2, t_n = 3t_n−1 for n ≥ 2.
• 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.Answer: (i) 80 × 0.6⁵ = 6.2208 m (ii) 80 + 2 × (48 + 28.8 + 17.28 + 10.368 + 6.2208) = 80 + 221.3376 = 301.3376 m (≈ 301.34 m).
• Which term of the sequence 2, 2√2, 4, … is 128?Answer: It is a GP with r = √2: 2 × (√2)ⁿ⁻¹ = 128 ⇒ (√2)ⁿ⁻¹ = 64 = (√2)¹² ⇒ n = 13. 128 is the 13th term.
• The figure 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 the figure and try to answer the following questions.Answer: (i) 1, 8, 64, 512 (ii) 4096 and 32768 (iii) t_n = 8ⁿ; t₀ = 1, t_n = 8t_n−1 for n ≥ 1 (iv) 89, 6481, 512729; 40966561, 3276859049; s_n = (89)ⁿ; s₀ = 1, s_n = 89s_n−1; the area gets closer and closer to 0.
• Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.Answer: d = 7, a = −32, so the 31st term = −32 + 30 × 7 = 178.
• Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.Answer: d = 6, a = 4: the AP is 4, 10, 16, 22, 28, …
• 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.)Answer: 105, 112, …, 994 is an AP with d = 7: 105 + (n − 1)7 = 994 gives n = 128.
• 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.)Answer: 12, 16, …, 248 is an AP with d = 4: 12 + (n − 1)4 = 248 gives n = 60.
• Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.Answer: r² = 4, so r = 2 or −2. r = 2: −43, −83, −163, …; r = −2: 4, −8, 16, −32, … Both GPs work.
• Find all possible ways of expressing 100 as the sum of consecutive natural numbers.Answer: 18 + 19 + 20 + 21 + 22 = 100 and 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100. These are the only ways with two or more numbers (apart from the single number 100 itself).
• 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?Answer: End of 2nd hour: 120; end of 4th hour: 480; end of nth hour: 30 × 2ⁿ.
• 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.Answer: d = 5, a = −13: the first three terms are −13, −8, −3.
• Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.Answer: S₄₄ = 990 ≤ 1000 and S₄₅ = 1035 > 1000, so the smallest value is n = 45.
• Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.Answer: 131072 is the 9th term. Explicit: t_n = 2 × 4ⁿ⁻¹. Recursive: t₁ = 2, t_n = 4t_n−1 for n ≥ 2.
• The sum of the first three terms of a GP is 1312 and their product is −1. Find the common ratio and the terms.Answer: r = −34 with terms 43, −1, 34, or r = −43 with terms 34, −1, 43.
• If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.Answer: x = ar³, y = ar⁹, z = ar¹⁵, so yx = zy = r⁶. The ratio is the same, so x, y, z are in GP (equivalently y² = xz).
• 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.Answer: The terms are 2, 6, 18 (r = 3), or the same numbers in the order 18, 6, 2 (r = 13).
• 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?Answer: 1, 2, 4, 8, 16, 32, 64, 128. Simpler rule: P_n = 2P_n−1 (for n ≥ 2, with P₁ = 1). Explicit: P_n = 2ⁿ⁻¹.
• 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?Answer: 1, 2, 3, 5, 8, 13, 21, 34. It is the Virahānka–Fibonacci sequence: each term is the sum of the two before it.

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