Class 9 Maths Chapter 8 – Predicting What Comes Next: Exploring Sequences and Progressions

Learning Objectives

After studying this chapter you should be able to:

• Recognise a sequence as an ordered list of numbers and name each number its term, using the notation \(t_1, t_2, t_3, \ldots\) (and \(s_n, u_n\) for other sequences).
• Describe sequences that are finite or infinite, increasing or decreasing, and those made of fractions or negative numbers.
• Write an explicit rule \(t_n = \text{(expression in } n)\) and use it to find any specified term directly, without listing earlier terms.
• Use an explicit rule to test whether a number is a term and to find its position in the sequence.
• Write a recursive rule from a starting value and a relation between terms, and generate successive terms from it.
• Recognise recurrences that use two or more previous terms, including the Virahānka–Fibonacci sequence.
• Identify an arithmetic progression, state its common difference \(d\), and write its \(n\)th term \(t_n = a + (n-1)d\) and recursive rule.
• Visualise an AP as points lying on a straight line, and handle fractional, decimal and negative common differences.
• Derive and apply the formula \(S_n = \dfrac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers, and connect it to triangular numbers and consecutive sums.
• Identify a geometric progression, state its common ratio \(r\), and write its \(n\)th term \(t_n = ar^{n-1}\) and recursive rule.
• Analyse the Sierpiński triangle and other fractals: count the black triangles \(3^n\) and follow the shrinking area \(\left(\tfrac{3}{4}\right)^n\) using the stage index starting at 0.
• Compare AP and GP growth, read graphs of sequence points, and apply sequences to realistic problems such as fares, seating, salaries and bouncing balls.

1. Patterns All Around Us

Patterns surround us. The rings on a cross-section of a tree, the petals of a flower, the beats of a song, the rows in a stadium, the interest your savings earn, the tiling on a floor — all are built from patterns. Patterns help us make sense of the world and, crucially, they help us predict what comes next. In mathematics, a sequence is a special kind of pattern: numbers or other objects arranged in a particular order. By studying sequences we can explore how numbers grow, shrink or repeat, and we can even use these ideas to solve real-life problems.

The three dots \( \ldots \) tell us that the sequence continues indefinitely. You have already met all four of these in Grades 6, 7 and 8; in this chapter we give them a precise language so that we can describe any sequence — even a strange or unexpected one — with a rule.

2. What Is a Sequence?

In the sequence of square numbers, \(1\) is the first term, \(4\) is the second term, \(25\) is the fifth term, and so on. A sequence may be finite or infinite. The four sequences above are infinite — they never end. But \(6, 12, 24, 48, 96\) is a finite sequence of exactly five terms. Can you think of finite sequences around you? The heights of the steps in a staircase, and the marks of the home matches of a cricket team in a season, are both examples of finite sequences.

To talk about a sequence conveniently, we need a notation. We use

The notation connects the position of a term to the term itself: \(t_4 = 7\) says that the term in the fourth position is \(7\). Often we need to discuss more than one sequence at a time, so we may use different letters: \(t_1, t_2, t_3, \ldots\) for one sequence, \(s_1, s_2, s_3, \ldots\) for another, and \(u_1, u_2, u_3, \ldots\) for a third.

Sequences come in many shapes. Their terms may be fractions or negative numbers, and they may increase or decrease.

3. Triangular Numbers

Look at the sequence \(1, 3, 6, 10, 15, 21, \ldots\). The differences between consecutive terms (among the first six terms) are \(2, 3, 4, 5\) and \(6\). But there is a much deeper way to read this sequence: each term can be rewritten as a sum of the natural numbers from 1 up to that position:

So each triangular number is the sum of the natural numbers up to that term. For example, \(15\), the fifth triangular number, equals \(1 + 2 + 3 + 4 + 5\). The name comes from the fact that these numbers can be drawn as triangular arrays of dots, as in the figure below.

Can you draw the dot patterns for the next two terms of the sequence? The sixth triangular number is \(21 = 1 + 2 + 3 + 4 + 5 + 6\), and the seventh is \(28\).

4. Square Numbers and Odd Numbers

Now look at the square numbers \(1, 4, 9, 16, 25, 36, \ldots\). The differences between consecutive terms (for the first six terms) are \(3, 5, 7, 9\) and \(11\) — which are precisely the odd numbers. There is an even more striking connection: every square number is a sum of the odd numbers up to that term:

Geometrically, when we build a larger square from a smaller one we add a bent row (an "L-shape") along two sides. That L-shape always contains exactly the next odd number of dots: from \(1\) to \(4\) we add \(3\) dots; from \(4\) to \(9\) we add \(5\) dots; from \(9\) to \(16\) we add \(7\) dots, and so on.

This relationship between the odd numbers and the square numbers is the reason the odd numbers deserve their own sequence: \(1, 3, 5, 7, 9, 11, \ldots\). You may recall some of these ideas from Grade 6, Chapter 1!

5. From Pattern to Rule

Consider the sequence \(1, 4, 7, 10, 13, \ldots\). What comes next? Since each term is 3 more than the previous one, the next terms are \(16, 19, 22, 25, \ldots\). If we give each term a position number, the sequence becomes a table of position against value:

The pattern "each term is 3 more than the previous one" becomes the rule for finding any term we want. But we can push further. The chapter asks: what do we get if we add all the terms up to a given term? The hint is to add cumulatively:

The cumulative sums are \(1, 5, 12, 22, 35, \ldots\) — a completely new sequence (the differences are \(4, 7, 10, 13, \ldots\)). This shows that studying a sequence often leads to interesting new ones.

6. Section 8.2 — Explicit Rule for a Sequence

Using the notation \(t_n\) (or \(s_n\) or \(u_n\)) we can write a rule for the term in the \(n\)th position — that is, the \(n\)th term.

7. Checking Whether a Number Is a Term

The explicit rule is useful in another way too: it can tell us whether a given number is a term of the sequence, and if so, which position it occupies. The idea is to reverse the rule: put the number into the formula and solve for \(n\). If the resulting \(n\) is an allowed position number (a positive integer such as \(1, 2, 3, \ldots\)), the number is a term; if not, it is not a term.

8. More Explicit Formulas

Simple patterns often lead to simple explicit formulas. Here we practise the two directions: given a position, find the value, and given a value, find the position (or decide whether it occurs).

9. The Sequence of Prime Numbers

Here is the sequence of the first ten prime numbers:

Do you see a pattern in this sequence? The differences are \(1, 2, 2, 4, 2, 4, 2, 4, 6, \ldots\) — not constant, and not following an obvious rhythm. This is a sequence where discovering a simple rule that lets an elementary school student predict the next terms is difficult. The primes remind us that not every sequence obeys an obvious rule. There is a big difference between:

• sequences with a simple known rule (the odd numbers, the squares, the triangular numbers), and
• sequences where finding a useful rule is genuinely difficult even though the pattern is "really there" (the primes).

10. Section 8.3 — Recursive Rule for a Sequence

So far we have described every sequence with an explicit formula in terms of \(n\). There is another way: a sequence can be described by relating each term to earlier terms.

Consider again the sequence \(1, 4, 7, 10, 13, \ldots\). Its explicit rule is \(t_n = 3n - 2\) (check that \(t_1 = 1\) and \(t_2 = 4\), and so on). But we also notice that every term is 3 more than the previous term:

To use a recursive rule, you must first know the earlier terms. That is the trade-off: recursive rules are brilliant at building the sequence one term at a time, but if you want the 100th term, you must generate all the terms before it.

11. Recursive Rules: More Examples

12. Recursive Rules Using Two or More Previous Terms

A recursive rule does not have to involve only the previous term — it may involve the previous two or more terms.

Notice that three starting values were needed before we could begin. The more previous terms a rule uses, the more starting values it needs.

13. The Virahānka–Fibonacci Sequence

The most famous two-term recurrence in all of mathematics begins like this:

Each new term is the sum of the previous two terms. Let us compute the next few:

\(V_3 = V_2 + V_1 = 2 + 1 = 3\)
\(V_4 = V_3 + V_2 = 3 + 2 = 5\)
\(V_5 = V_4 + V_3 = 5 + 3 = 8\)

So the sequence begins

and each term after the first two is obtained by adding the previous two. Can you write the next two terms? They are \(34 + 21 = 55\) and \(55 + 34 = 89\).

Do you recognise this sequence from earlier grades? It is the Virahānka–Fibonacci sequence. The source chapter tells its remarkable story:

The Virahānka sequence plays an important role throughout mathematics and science — in nature (spirals, branching), in art, and in later chapters of your own mathematics education. We will encounter it again many times.

14. Explicit vs Recursive Rules: A Comparison

Both kinds of rule can describe the same sequence. Take \(3, 7, 11, 15, \ldots\):

15. Section 8.4 — Arithmetic Progressions

Among all sequences, one special family stands out: the sequences in which the difference between any two consecutive terms is constant.

You have already met several APs in this chapter!

Why is \(-2\) a valid common difference? Because \(9 - 7 = 2\) disguised with a sign: we must compute \(t_2 - t_1\) in the same direction as the sequence. For \(9, 7, 5, \ldots\), \(t_2 - t_1 = 7 - 9 = -2\). So:

• \(d > 0\) → the AP increases.
• \(d < 0\) → the AP decreases.
• \(d = 0\) → the AP is constant (all terms equal).

16. Identifying an AP

To check whether a given sequence is an AP, we simply compute the differences of consecutive terms and see whether they are all equal.

17. Section 8.4.1 — Visualising an AP

Because the common difference is constant, the terms of an AP are equally spaced: on a number line, they sit at uniform gaps exactly like a staircase. Consider the AP \(1, 5, 9, 13, \ldots\) with \(d = 4\). If we build a small grid and add a new square on each stage, the number of squares at each stage forms exactly this AP:

The most important visual property of an AP, however, shows up on a graph. Plot each pair (stage number, number of squares) as a point — \((1, 1)\), \((2, 5)\), \((3, 9)\), \((4, 13)\), \((5, 17)\):

The points fall on a straight (dashed) line. This happens precisely because the differences are constant: adding \(d\) to the term is like climbing the same step each time. This straight-line property is a handy way to spot an AP in a graph, and we will build graphs just like this again when we study linear equations.

18. APs with Fractions and Decimals

An AP does not need whole-number terms. The only requirement is a constant difference.

The last one is decreasing because its common difference is negative.

19. AP Recursive Rule and an Application

An AP has an especially simple recursive rule: start at \(a\), then keep adding \(d\).

The explicit rule says: start at \(a\), then add \(d\) once for every step after the first. Since there are \(n-1\) such steps, we add \((n-1)d\). (Check \(n = 1\): \(t_1 = a + 0 = a\).)

20. Section 8.5 — Sum of the First \(n\) Natural Numbers

Now we try something new: adding the terms of a simple sequence. How much is

Let us record the partial sums of the natural numbers as we go:

The partial sums \(1, 3, 6, 10, 15, \ldots\) — those are exactly the triangular numbers again! And for \(n = 10\) we get 55.

But what if we need \(1 + 2 + \cdots + 100\)? Listing and adding 100 numbers is tedious. There is a wonderful short-cut rule.

Let us check with \(n = 10\): \(S_{10} = \dfrac{10 \times 11}{2} = 55\). It works! And for \(n = 100\): \(S_{100} = \dfrac{100 \times 101}{2} = 5050\). One line of arithmetic instead of a hundred additions.

21. A Visual Proof of the Natural-Number Sum

Why should \(1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}\)? There is a beautiful picture. Take a staircase that has \(1, 2, 3, \ldots, n\) dots in its steps. Two copies of that staircase can interlock into a rectangle of size \(n\) by \(n+1\).

22. A Historical Note — Āryabhaṭa and the Āryabhaṭīya

Who first found this neat rule for adding natural numbers? The chapter carries a fascinating piece of history:

Such a "sum of a series" formula is immensely useful in practical computation — from counting objects arranged in triangular piles to rapid bookkeeping. We will build on it further in Section 24.

23. Sums of Consecutive Integers

The rule \(S_n = \dfrac{n(n+1)}{2}\) lets us add the first \(n\) natural numbers in one shot. What if we want a block of consecutive integers that does not start at 1, like \(25 + 26 + \cdots + 58\)? We already saw the trick in Example 15: compute the full sum up to the last number, then subtract the full sum up to the number just before the block starts. Any block of consecutive integers is therefore just a difference of two "triangular" sums.

24. The Triangular-Number Connection

We have now found the same numbers twice. The sum of the first \(n\) natural numbers produces

\[S_n = 1 + 2 + \cdots + n = \frac{n(n+1)}{2}\]

and the \(n\)th triangular number (Section 3) was exactly the dot pattern \(1, 3, 6, 10, 15, \ldots\). So the two are identical:

This is a beautiful example of how one idea in mathematics (adding natural numbers) and another quite different-looking idea (counting dots in a triangle) turn out to be the same idea.

25. Section 8.6 — Geometric Progressions

An AP grows by adding the same number each time. What if a sequence grows by multiplying by the same number each time?

Note that multiplying by \(\frac{1}{3}\) makes the terms shrink, and multiplying by a negative ratio makes the terms alternate sign. Just as \(d\) is found by subtraction (\(t_{n} - t_{n-1}\)), the common ratio is found by division: \(r = \dfrac{t_{n}}{t_{n-1}}\).

26. Identifying a GP

To check whether a sequence is a GP, compute the ratios of consecutive terms and see whether they are all equal.

27. GPs with Fractions and Decimals

Just like APs, GPs may have fractional or decimal terms.

In the last sequence the terms are positive and negative alternately because the ratio itself is negative. A ratio between 0 and 1 shrinks the terms; a ratio greater than 1 grows them; a ratio equal to 1 makes the GP constant.

28. GP Recursive Rule

Just as an AP has a recursive description, so does a GP: start at \(a\), then keep multiplying by \(r\).

The building formula says: start with \(a\), then multiply by \(r\) once for every step after the first, giving \(n-1\) multiplications in all. For the GP \(3, 6, 12, 24, \ldots\) we get \(t_5 = 3 \times 2^4 = 3 \times 16 = 48\).

29. Section 8.6.1 — Fun with Fractals: The Sierpiński Triangle

Here is a pattern that keeps repeating itself at smaller and smaller scales — a fractal. Start with one equilateral triangle (Stage 0). Divide it into four smaller triangles and cut out the middle one; the remaining picture is Stage 1. Do the same to each remaining black triangle to get Stage 2, Stage 3, and so on. This object is called the Sierpiński triangle, after the Polish mathematician Wacław Sierpiński (1882–1969).

The numbers of black triangles form a sequence: \(1, 3, 9, 27, \ldots\). Going from one stage to the next, the number of black triangles is multiplied by 3 — this is a geometric progression! (You may also recall a similar shape from Grade 6, Chapter 11.)

30. Sierpiński Triangle: Counting the Black Triangles

Let us write down the counts stage by stage:

Each stage multiplies the count by 3, so the count at stage \(n\) is \(3^n\). Note that the story starts at stage 0, which is why the count is \(3^0 = 1\). Recursively, the rule is:

31. Sierpiński Triangle: How Much Black Remains?

There is a second sequence hiding in the Sierpiński triangle. If the whole Stage 0 triangle has area 1 (one whole unit), then at Stage 1 we remove the middle sub-triangle, whose area is \(\frac{1}{4}\) of the whole. The black area becomes \(1 - \frac{1}{4} = \frac{3}{4}\). At every later stage we remove \(\frac{1}{4}\) of each remaining black triangle, so the black area is multiplied by \(\frac{3}{4}\) each time:

This is a GP with \(r = \frac{3}{4}\) — a shrinking GP. The black area keeps getting smaller and smaller, approaching 0 as the stage number grows, although every single stage still contains some black.

32. Fractals in Nature

The Sierpiński triangle is a mathematical fractal: a shape built from copies of itself at ever smaller scales. Surprisingly, nature is full of such self-similar patterns:

• Branches and veins: each branch of a tree splits into smaller branches that look like the whole tree; leaves have veins that split like the leaf itself; rivers branch into streams, which branch into rivulets.
• Cauliflower and fern: the florets of some vegetables (such as romanesco) spiral in smaller copies of the whole vegetable; fern leaves repeat the leaf shape in miniature.
• Clouds, snowflakes and coastlines: a cloud looks similar from close up or far away, and a coastline shows the same jumble of bays and capes at many magnifications.
• Lungs and ferns in us: the branching of the airways in our lungs packs a huge surface into a small space — nature uses self-similarity to be efficient.

All of these are rough, irregular versions of the idea you just met in the perfect, regular Sierpiński triangle. Mathematics gives us a language for how nature builds them through repeated multiplication — our geometric progressions.

33. Section 8.6.2 — Visualising a GP

Just as we visualised an AP with a grid and a graph, let us do the same for a GP. Stage 1 has 3 squares; each new stage has twice as many squares as the previous one.

Now plot the points (stage, number of squares): \((1, 3)\), \((2, 6)\), \((3, 12)\), \((4, 24)\), \((5, 48)\).

Contrast this with the AP graph in Section 17, which was a straight line. That difference — straight line for adding, curving fast for multiplying — is one of the most useful ideas in the whole chapter.

34. Application: The Bouncing Ball

Here is a classic application of a shrinking GP, taken from Section 8.6.2.

35. AP vs GP: Comparing Growth

Arithmetic and geometric progressions are the two giant families of sequences. Here is how they compare side by side.

36. Key Definitions

37. Formula Bank

38. Solved Examples

The worked examples below cover every type of question in this chapter. Each one is the kind you will face in exercises and exams. (All are original or drawn from the NCERT exercise sets; none are fabricated board questions.)

39. Application-Based Examples

40. Competency-Based Examples

These questions ask you to think beyond the formula: interpret, compare, justify, and connect ideas.

41. Case-Based Examples

42. Common Mistakes to Avoid

43. Exam Focus

Based on this chapter, boards and school exams typically ask:

• Very short answer: state whether an AP/GP is given, write the next term, or give \(a\), \(d\), \(r\).
• Short answer: find the \(n\)th term; test whether a number is a term; write a recursive or explicit rule from a pattern.
• Medium answer: sums of natural numbers and triangular numbers; block sums like \(25 + 26 + \cdots + 58\); find \(a\) and \(d\) (or \(a\) and \(r\)) from two terms.
• Application: taxi fares, seating, savings, bouncing balls, bacteria, Sierpiński counts and areas.
• Reasoning: compare AP vs GP; justify "not a term"; interpret graphs (straight line = AP, curved = GP).

44. Quick Revision

45. Self-Practice Questions

1. Write the first five terms of the sequence with \(t_n = 4n - 3\).
2. For \(t_n = 5n + 2\), find \(t_{20}\).
3. Which term of \(3, 8, 13, 18, \ldots\) is 3? Which term is 98? (Hint: use \(t_n = 3n -? \))
4. Is 2021 a term of the sequence \(1, 5, 9, 13, \ldots\)? Justify.
5. Using \(u_1 = 2\), \(u_{n+1} = u_n + 5\), write the first six terms.
6. Using \(v_1 = 1\), \(v_2 = 4\), \(v_n = v_{n-1} + v_{n-2}\), write the first six terms.
7. Write the first ten terms of the Virahānka–Fibonacci sequence and give \(V_9\) and \(V_{10}\).
8. An AP has \(a = 3\) and \(d = 7\). Find its 12th term.
9. An AP has \(t_4 = 19\) and \(t_9 = 44\). Find \(a\) and \(d\).
10. In the AP \(5, 11, 17, \ldots\), which term is 95?
11. Which of \(1, 2, 4, 8\); \(1, 4, 9, 16\); \(1, 5, 9, 13\); \(1, 1, 1, 1\) are APs? Which are GPs?
12. The AP \(\frac{1}{2}, 1, \frac{3}{2}, \ldots\) — find its 10th term.
13. Write the first five terms of the AP with \(a = -8\) and \(d = 3\).
14. Give an explicit rule for the AP \(7, 4, 1, -2, \ldots\).
15. Find \(1 + 2 + \cdots + 20\).
16. Find \(13 + 14 + \cdots + 20\) using \(S_{20} - S_{12}\).
17. Write the 12th and the 20th triangular numbers.
18. Find \(1 + 2 + \cdots + 99\).
19. If \(1 + 2 + \cdots + n = 55\), find \(n\).
20. If \(1 + 2 + \cdots + n = 105\), find \(n\).
21. A GP has \(a = 3\), \(r = 4\). Find its 6th term.
22. For the GP \(2, 6, 18, \ldots\), which term is 486?
23. For the GP \(16, 8, 4, \ldots\), find its 8th term.
24. For the GP \(1, -2, 4, -8, \ldots\), find its 7th term.
25. For the GP \(2, -6, 18, \ldots\), find its 5th term.
26. Write the first five terms of the GP with \(a = 1\), \(r = \frac{1}{2}\).
27. A GP has \(a = 1\), \(r = 3\). Find its 7th term.
28. In the Sierpiński triangle, how many black triangles are at stages 5 and 6?
29. Write the Sierpiński black area at stages 2 and 3 (leave as fractions).
30. At which stage does the Sierpiński black area first fall below half of the original?
31. A taxi ride charges a flag-down of ₹250 plus ₹45 per km. What is the fare for an 11 km ride?
32. A stadium has 20 seats in row 1 and 4 extra per row. How many seats are in row 25?
33. Ravi saves ₹100 in week 1, increasing by ₹50 weekly. How much does he save in week 20?
34. A ball dropped from 40 m rebounds to \(\frac{3}{4}\) of its height. What is the height after the 3rd rebound?
35. A colony of 100 bacteria doubles hourly. How many are there after 4 hours?
36. Give an example of a sequence that is BOTH an AP and a GP.
37. True or false: the graph of an AP is a straight line; the graph of a GP is also straight.
38. Explain, in one sentence, why the sum of the first \(n\) natural numbers is the same as the \(n\)th triangular number.
39. Classify each: \(2, 5, 8\);   \(2, 6, 18\);   \(1, \frac{1}{2}, \frac{1}{4}\);   \(1, 2, 3, 5\).
40. Challenge: the 5th term of a GP is 48 and the common ratio is 2. Find the first term.

46. Self-Practice Answers

1. \(1, 5, 9, 13, 17\).
2. \(102\).
3. \(3\) is the 1st term. The AP is \(t_n = 5n - 2\); \(98 = 5n - 2 \Rightarrow n = 20\), so 98 is the 20th term.
4. Yes: \(4n - 3 = 2021 \Rightarrow 4n = 2024 \Rightarrow n = 506\), a whole number, so 2021 is the 506th term.
5. \(2, 7, 12, 17, 22, 27\).
6. \(1, 4, 5, 9, 14, 23\).
7. \(1, 2, 3, 5, 8, 13, 21, 34, 55, 89\);   \(V_9 = 55\), \(V_{10} = 89\).
8. \(3 + 11 \times 7 = 80\).
9. \(a + 3d = 19\), \(a + 8d = 44 \Rightarrow d = 5\), \(a = 4\).
10. \(5 + 6(n-1) = 95 \Rightarrow n = 16\).
11. APs: \(1, 5, 9, 13\) and \(1, 1, 1, 1\). GP: \(1, 2, 4, 8\) (and \(1, 1, 1, 1\) also has ratio 1). \(1, 4, 9, 16\) is neither.
12. \(t_{10} = \frac{1}{2} + 9 \times \frac{1}{2} = 5\).
13. \(-8, -5, -2, 1, 4\).
14. \(t_n = 10 - 3n\).
15. \(\frac{20 \times 21}{2} = 210\).
16. \(S_{20} - S_{12} = 210 - 78 = 132\).
17. \(T_{12} = 78\), \(T_{20} = 210\).
18. \(\frac{99 \times 100}{2} = 4950\).
19. \(\frac{n(n+1)}{2} = 55 \Rightarrow n = 10\).
20. \(\frac{n(n+1)}{2} = 105 \Rightarrow n(n+1) = 210 \Rightarrow n = 14\).
21. \(3 \times 4^5 = 3072\).
22. \(2 \times 3^{n-1} = 486 \Rightarrow 3^{n-1} = 243 = 3^5 \Rightarrow n = 6\).
23. \(16 \times \left(\frac{1}{2}\right)^{7} = \frac{1}{8}\).
24. \(1 \times (-2)^6 = 64\).
25. \(r = -3\); \(2 \times (-3)^4 = 162\).
26. \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\).
27. \(3^6 = 729\).
28. Stage 5: \(3^5 = 243\); stage 6: \(3^6 = 729\).
29. Stage 2: \(\left(\frac{3}{4}\right)^2 = \frac{9}{16}\); stage 3: \(\frac{27}{64}\).
30. Stage 3 (since \(\frac{9}{16} = 0.5625 > \frac{1}{2}\) but \(\frac{27}{64} \approx 0.42 < \frac{1}{2}\)).
31. ₹\(250 + 45 \times 11 =\) ₹745.
32. \(16 + 4 \times 25 = 116\) seats.
33. \(50 + 50 \times 20 =\) ₹1050.
34. \(40 \times \left(\frac{3}{4}\right)^3 = 40 \times \frac{27}{64} = 16.875\) m.
35. \(100 \times 2^4 = 1600\).
36. Any constant sequence, e.g. \(7, 7, 7, \ldots\) (has \(d = 0\) and \(r = 1\)).
37. True for the AP (straight line); false for the GP (it curves).
38. Because the nth triangular array has exactly \(1 + 2 + \cdots + n\) dots.
39. \(2, 5, 8\) is an AP; \(2, 6, 18\) is a GP; \(1, \frac{1}{2}, \frac{1}{4}\) is a shrinking GP; \(1, 2, 3, 5\) is the start of the Virahānka–Fibonacci-like sequence.
40. \(a \times 2^4 = 48 \Rightarrow a = 3\).