NCERT Solutions for Class 9 Maths Chapter 8 Predicting What Comes Next Exploring Sequences and Progressions (PDF) – 2027

Each of the four sequences is described by what happens from one term to the next, and that alone fixes every later term.

Check the last two rows against the printed terms: 21 + 7 = 28, 28 + 8 = 36, 36 + 9 = 45; and 36 + 13 = 49, 49 + 15 = 64, 64 + 17 = 81. The gaps themselves form a sequence, and that is what makes prediction possible.

Yes — a finite sequence is any ordered list that stops, and everyday life is full of them.

• The overs bowled in a T20 innings: 1, 2, 3, …, 20.
• The roll numbers of the 40 students in a class: 1, 2, 3, …, 40.
• The marks of a student in six subjects, written in the order of the report card.
• The number of days in the twelve months of 2026: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.
• The stairs of a flight numbered from bottom to top, or the platform numbers at a railway station.

Fig. 8.1 shows the first five triangular numbers 1, 3, 6, 10, 15 as triangular arrays of dots. The next two terms are 21 and 28, drawn by adding one more row each time.

Fig. 8.2 fills a square array of dots with L-shaped bands (gnomons), each one a different colour. The first band is 1 dot, the second 3, the third 5, and so on — so building a square is the same as adding up odd numbers.

The gap between consecutive terms is a constant 3, so the next four terms are 16, 19, 22, 25.

Now build the running totals. Take the first ten terms 1, 4, 7, 10, 13, 16, 19, 22, 25, 28 and add them one at a time:

So the required sequence is 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

The triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, … , so

Sequences differ in what their terms are made of and in how one term leads to the next. Five clearly different types:

Because it lets you reach any term directly, without first working out all the terms before it.

It is also the only convenient way to answer the reverse question. To ask “is 137 in this sequence, and where?” you simply solve the equation:

The sequence is 1, 4, 9, 16, 25, 36, … , and the rule is

Test it against the printed terms: t₁ = 1² = 1, t₂ = 2² = 4, t₃ = 9, t₄ = 16, t₅ = 25, t₆ = 36. All six agree.

Substitute the position straight into the rule.

Why n must be a natural number: n is the position of a term (1st, 2nd, 3rd, …). There is no 94.6th place, so if s_n = k gives an n that is not a natural number, k is not a term.

There are visible features, but no rule that predicts the next term. Compare the gaps:

The gaps do not settle to a constant (so this is not an AP), and the ratios 3/2, 5/3, 7/5, … are not constant either (so it is not a GP). The next few primes — 31, 37, 41, 43, 47 — have to be found by testing for divisibility, not produced by a formula.

(i) Substitute each position into t_n = 3n – 7.

(ii) Solve t_n = 332.

So 332 is the 113th term.

(iii) Solve t_n = 557 the same way.

188 is a natural number, so yes — 557 is a term of the sequence, the 188th term. Check: 3 × 188 – 7 = 564 – 7 = 557.

First four terms (as in the book): u₁ = 1, u₂ = 5, u₃ = 13, u₄ = 29. Continue with u_n = 2u_n–1 + 3:

The terms keep increasing: 1, 5, 13, 29, 61, 125, 253, … and 133 lies between u₆ = 125 and u₇ = 253. So 133 is not a term of this sequence.

Use the recursive rule V_n = V_n–1 + V_n–2, adding the last two terms each time.

The sequence continues 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

(i) t_n = 3n – 4

(ii) t_n = 2 – 5n

(iii) t_n = n² – 2n + 3

Solve t_n = 97 and t_n = 172 and see whether n comes out a natural number.

Both values of n are natural numbers, so 97 is the 20th term and 172 is the 35th term — both are terms of the sequence.

So 607 is the 122nd term. Check: 5 × 122 – 3 = 610 – 3 = 607. ✓

Start at t₁ = –5 and add 3 each time.

The terms rise by a constant 3, so this is an AP with a = –5 and d = 3. Convert to an explicit rule and then solve:

Yes — 52 is a term, the 20th term. Check: 3 × 20 – 8 = 52. ✓

Each term is the sum of the three terms before it.

The sequence is 1, 2, 4, 7, 13, 24, 44, 81, …

Yes — and once the rule is justified, every one of these follows by substitution. Each stage adds 4 squares, one to each arm of the X, so the counts are 1, 5, 9, 13, 17, 21, …

The test is single and mechanical: subtract each term from the one after it. If every one of those differences is the same number, the sequence is an AP; one exception is enough to rule it out.

(i) 2, 5, 8, 11, … — differences 5 – 2 = 3, 8 – 5 = 3, 11 – 8 = 3. Constant, so it is an AP with a = 2, d = 3.

(ii) –5, –1, 3, 7, … — differences –1 – (–5) = 4, 3 – (–1) = 4, 7 – 3 = 4. Constant, so it is an AP with a = –5, d = 4.

What the plot shows: plotting (1, 2), (2, 5), (3, 8), (4, 11) — and likewise (1, –5), (2, –1), (3, 3), (4, 7) — the points lie exactly on a straight line, just as in Fig. 8.4.

(i) Here a = 1/2 and d = 5/2 – 1/2 = 4/2 = 2 (also 9/2 – 5/2 = 2 and 13/2 – 9/2 = 2).

(ii) Here a = 1.5 and d = 3.5 – 1.5 = 2.

A recursive rule for an AP needs two things: the first term, and the instruction “add d”.

The total fare for 10 km is ₹600 (₹200 booking fee + 10 × ₹40).

Yes — the method does not care how long the list is. Write the sum forwards, write it backwards underneath, and add column by column.

Yes. Substitute into S_n = n(n + 1)/2.

Check the first of these directly: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55. ✓

First term a = 3; common difference d = 8 – 3 = 5 (and 13 – 8 = 5, 18 – 13 = 5).

Here a = 21 and d = 18 – 21 = –3, so the AP falls.

So –81 is the 35th term. Check: 24 – 3 × 35 = 24 – 105 = –81. ✓

Is 0 a term? Solve 24 – 3n = 0.

Yes — 0 is a term, the 8th term, because n = 8 is a natural number. Listing confirms it: 21, 18, 15, 12, 9, 6, 3, 0, –3, …

a = 11 and d = 8 – 11 = –3 (also 5 – 8 = –3, 2 – 5 = –3).

Recursive rule:

Write both facts using t_n = a + (n – 1)d. The last term is the 50th, so n = 50 gives a + 49d.

Subtract (1) from (2) — this eliminates a:

Now the 29th term:

The 2-digit multiples of 3 run from 12 to 99, and they form an AP with d = 3.

So there are 30 such numbers. For the sum, pair the ends as in Section 8.5:

His annual salaries form an AP with a = ₹5,00,000 and d = ₹20,000.

After 10 years (that is, in his 11th year of work) his income reaches ₹7,00,000.

The total is 1 + 2 + 3 + … + 25 — the sum of the first 25 natural numbers.

Each stage of Fig. 8.6 is a rectangle 3 squares wide whose height doubles: 1, 2, 4, 8 rows. So the counts double rather than grow by a fixed amount.

How it differs from Fig. 8.3:

In each sequence every term divided by the term before it gives the same number, so all three are GPs, with common ratios 2, 3 and –1 respectively.

(i) 2, 10, 50, 250, …

(ii) 4, 8/3, 16/9, 32/27, …

(iii) 3, –3/2, 3/4, –3/8, …

Each term is ten times the one before, so:

Running it: t₂ = 10 × 3 = 30, t₃ = 10 × 30 = 300, t₄ = 10 × 300 = 3000. ✓

At each new stage every black triangle becomes 3 black triangles, and 3/4 of the black area is kept.

(a) 1, 3, 9, 27   (b) Stage 4: 81; Stage 5: 243

(c) t_n = 3ⁿ (recursively: t₀ = 1, t_n = 3 × t_n–1)

(d) Stages 1, 2, 3: 3/4, 9/16, 27/64; Stages 4, 5: 81/256, 243/1024 square units. s_n = (3/4)ⁿ (recursively: s₀ = 1, s_n = (3/4) × s_n–1). As n increases, the black area gets closer and closer to 0, but never becomes 0.

Because the growth is by multiplication, not addition. Every black triangle turns into three at the next stage, so the whole count triples each time.

Compare this with the AP of Fig. 8.3, where each stage adds a fixed 4:

Because the same operation is applied to every black triangle at every stage, and it always keeps exactly three of four equal parts.

Applying this repeatedly from Stage 0, where the area is 1:

The 12th term is four steps beyond the 8th, and each step multiplies by r = 2.

a = 5 and r = 25 ÷ 5 = 5 (also 125 ÷ 25 = 5).

Check the rule on the printed terms: t₁ = 5¹ = 5, t₂ = 5² = 25, t₃ = 5³ = 125. ✓

Generate the terms until 730 appears.

So 730 is the 7th term.

a = 2 and r = 6 ÷ 2 = 3 (also 18 ÷ 6 = 3).

Now solve t_n = 4374:

So 4374 is the 8th term. Check: 2 × 3⁷ = 2 × 2187 = 4374. ✓

The peak heights form a GP with r = 60% = 0.6, starting from the 80 m drop.

(i) The height after the 5th bounce is 6.2208 m (about 6.22 m).

(ii) The ball first falls 80 m. After each of the 5 bounces it rises and then falls the same height; the fall after the 5th bounce ends with the 6th hit.

First check that this is a GP by taking ratios.

Write every term as a power of 2, using √2 = 2^1/2:

128 is the 13th term. Check: t₁₃ = 2 × (√2)¹² = 2 × 2⁶ = 2 × 64 = 128. ✓

Each red square is cut into 9 equal squares and 8 of them are kept. So each stage multiplies the number of red squares by 8 and the red area by 8/9.

(i) 1, 8, 64, 512   (ii) Stage 4: 4096; Stage 5: 32768

(iii) Explicit: t_n = 8ⁿ; recursive: t₀ = 1, t_n = 8 × t_n–1 for n ≥ 1.

(iv) Stages 1, 2, 3: 8/9, 64/81, 512/729; Stages 4, 5: 4096/6561, 32768/59049 square units. Explicit: s_n = (8/9)ⁿ; recursive: s₀ = 1, s_n = (8/9) × s_n–1 for n ≥ 1. Since 8/9 < 1, the red area gets closer and closer to 0 as n increases, but never becomes 0.

Two terms of an AP give two equations in a and d.

“The 7th term exceeds the 5th by 12” means t₇ – t₅ = 12, and those two terms are two steps apart.

The AP is 4, 10, 16, 22, 28, 34, 40, …

Check: the 3rd term is 16 ✓; t₇ = 40 and t₅ = 28, and 40 – 28 = 12 ✓.

Find the two ends first.

These form an AP with a = 105, d = 7, last term 994.

There are 128 three-digit numbers divisible by 7.

The multiples of 4 strictly between 10 and 250 run from 12 to 248.

There are 60 multiples of 4 between 10 and 250.

Let the GP be a, ar, ar², … Translate both conditions.

Case r = 2: a(1 + 2) = –4 gives 3a = –4, a = –4/3.

Case r = –2: a(1 – 2) = –4 gives –a = –4, a = 4.

Both are valid answers: –4/3, –8/3, –16/3, … and 4, –8, 16, –32, …

Suppose the sum has k terms starting at a. Using the AP sum idea, the total is k terms whose average is (first + last)/2:

Now k and (2a + k – 1) are two factors of 200. They always have opposite parity: if k is odd, k – 1 is even so 2a + k – 1 is even, and vice versa. So one factor must be odd. The odd factors of 200 = 2³ × 5² are 1, 5 and 25.

Doubling every hour is a GP with a = 30 and r = 2, counting from hour 0.

Write each term as a + (n – 1)d and add.

The first three terms are –13, –8, –3.

Require S_n = n(n + 1)/2 > 1000, that is n(n + 1) > 2000.

So the smallest such value is n = 45, giving S₄₅ = 1035 > 1000, while S₄₄ = 990 < 1000.

a = 2 and r = 8 ÷ 2 = 4 (also 32 ÷ 8 = 4).

Solve t_n = 131072 by turning everything into powers of 2:

So 131072 is the 9th term. Check: 2 × 4⁸ = 2 × 65536 = 131072. ✓

Choosing the three terms as a/r, a, ar makes the product collapse.

Now use the sum:

Factorise: 12r² + 16r + 9r + 12 = 4r(3r + 4) + 3(3r + 4) = (3r + 4)(4r + 3).

With a = –1 and r = –3/4 the three terms are a/r = 4/3, a = –1, ar = 3/4:

The other root r = –4/3 gives the same three numbers in reverse order: 3/4, –1, 4/3.

Let the GP have first term a and common ratio r.

Compare the two ratios of consecutive members of x, y, z:

The ratio of consecutive terms is the same constant, so x, y, z form a GP with common ratio r⁶. Equivalently y² = a²r¹⁸ = (ar³)(ar¹⁵) = xz, which is the same statement.

Let the three terms be a, ar, ar².

Since 1 + r² + r⁴ = (1 + r²)² – r² = (1 + r + r²)(1 – r + r²), dividing (2) by (1) gives

r = 3 gives a = 2, so the terms are 2, 6, 18 (r = 1/3 gives the same numbers in reverse order: 18, 6, 2).

So the values are 1, 2, 4, 8, 16, 32, 64, 128.

Simpler recursive formula. For n ≥ 4, subtract the rule for P_n–1 from the rule for P_n:

This also holds for n = 3 (P₃ = 4 = 2P₂), so P₁ = 1, P_n = 2P_n–1 for n ≥ 2 — a GP with a = 1 and r = 2. Explicit formula: P_n = 2^n–1.

Note that this sum stops at W_n–2, not W_n–1 — it leaves out the term immediately before.

The sequence is 1, 2, 3, 5, 8, 13, 21, 34 — the Virahānka–Fibonacci sequence.