NCERT Solutions Class 9 Maths Exercise 8.1 — Predicting What Comes Next: Exploring Sequences and Progressions

NCERT Class 9 Mathematics — Ganita Manjari Part I, Predicting What Comes Next: Exploring Sequences and Progressions, Exercise Set 8.1. All 6 questions solved.

A sequence is an ordered list of numbers, and there are two quite different ways of describing one.

• An explicit rule gives the term from its position: put $n$ in, get $t_n$ out. To find $t_{100}$ you need nothing but the rule and the number $100$.
• A recursive rule gives each term from the ones before it, together with enough starting values to get going. To find $t_{100}$ you must climb all ninety-nine steps.

Key insight. An explicit rule turns “which term is $607$?” into an equation. Every question of that kind on this page — Questions 3, 4 and 5 — is solved by writing $t_n = $ the given value and solving for $n$, then checking that $n$ comes out a positive whole number. A fractional or negative $n$ is the proof that the value is not a term.

Question 1

Find the first five terms of the sequence in which the $n^{th}$ term is given by

Question 1 (i)

$t_n = 3n – 4$ for $n \ge 1$.

Solution. Substitute $n = 1, 2, 3, 4, 5$ in turn:

$$t_1 = 3(1) – 4 = -1, \qquad t_2 = 3(2) – 4 = 2, \qquad t_3 = 3(3) – 4 = 5$$ $$t_4 = 3(4) – 4 = 8, \qquad t_5 = 3(5) – 4 = 11$$

Notice that consecutive terms differ by $3$ throughout, which is no accident: the coefficient of $n$ in a linear rule is the common difference. So this is an AP with first term $-1$ and common difference $3$.

Question 1 (ii)

$t_n = 2 – 5n$ for $n \ge 1$.

Solution.

$$t_1 = 2 – 5 = -3, \qquad t_2 = 2 – 10 = -8, \qquad t_3 = 2 – 15 = -13$$ $$t_4 = 2 – 20 = -18, \qquad t_5 = 2 – 25 = -23$$

Here the coefficient of $n$ is $-5$, so the terms decrease by $5$ each time. The sign of that coefficient is the whole story of whether a linear sequence rises or falls.

Question 1 (iii)

$t_n = n^2 – 2n + 3$ for $n \ge 1$.

Solution.

$$t_1 = 1 – 2 + 3 = 2, \qquad t_2 = 4 – 4 + 3 = 3, \qquad t_3 = 9 – 6 + 3 = 6$$ $$t_4 = 16 – 8 + 3 = 11, \qquad t_5 = 25 – 10 + 3 = 18$$

This one is not an AP: the differences are $1, 3, 5, 7$ — themselves increasing by $2$. That pattern is the fingerprint of a quadratic rule, and it is worth recognising, because it tells you at a glance that no formula of the form $a + (n-1)d$ can produce this list.

Question 2

Find the $10^{th}$ and $15^{th}$ terms of the sequence $t_n = 5n – 3$ for $n \ge 1$.

Solution. This is what an explicit rule is for — no need to write out the earlier terms at all:

$$t_{10} = 5(10) – 3 = 50 – 3 = 47$$

$$t_{15} = 5(15) – 3 = 75 – 3 = 72$$

A check: from the $10$th term to the $15$th is five steps of $5$, so the terms should differ by $25$ — and $72 – 47 = 25$ ✓

Question 3

Determine whether $97$ and $172$ are terms of the sequence $t_n = 5n – 3$ for $n \ge 1$.

Solution. A number is a term exactly when the equation $5n – 3 = \text{that number}$ has a solution that is a positive integer.

Is $97$ a term?

$$5n – 3 = 97 \ \Longrightarrow \ 5n = 100 \ \Longrightarrow \ n = 20$$

$n = 20$ is a positive whole number, so yes — $97$ is the $20$th term.

Is $172$ a term?

$$5n – 3 = 172 \ \Longrightarrow \ 5n = 175 \ \Longrightarrow \ n = 35$$

Again a positive whole number, so yes — $172$ is the $35$th term.

A faster test. The sequence runs $2, 7, 12, 17, 22, \dots$, and every term leaves remainder $2$ on division by $5$. So a number is a term precisely when it is at least $2$ and leaves remainder $2$ when divided by $5$:

$$97 = 5 \times 19 + 2 \ \checkmark, \qquad 172 = 5 \times 34 + 2 \ \checkmark$$

By contrast $100$ leaves remainder $0$, so $100$ could never be a term however far the sequence ran — which is the kind of question this test settles instantly.

Question 4

Which term of the sequence $t_n = 5n – 3$ for $n \ge 1$ is $607$?

Solution.

$$5n – 3 = 607 \ \Longrightarrow \ 5n = 610 \ \Longrightarrow \ n = 122$$

Check by substituting back: $5(122) – 3 = 610 – 3 = 607$ ✓

Question 5

A sequence is given by the recursive rule $t_1 = -5$, $t_{n+1} = t_n + 3$ for $n \ge 1$. Find the first five terms of the sequence. Is $52$ a term of this sequence? If so, which term is it?

Solution.

The first five terms. Each term is $3$ more than the one before, starting at $-5$:

$$t_1 = -5$$ $$t_2 = -5 + 3 = -2$$ $$t_3 = -2 + 3 = 1$$ $$t_4 = 1 + 3 = 4$$ $$t_5 = 4 + 3 = 7$$

Is $52$ a term? Climbing one step at a time would work but would take twenty lines, so convert the recursive rule into an explicit one first. The rule “add $3$ each time” says this is an AP with $a = -5$ and $d = 3$:

$$t_n = a + (n-1)d = -5 + 3(n-1) = 3n – 8$$

Check it against the terms already found: $t_1 = 3 – 8 = -5$ ✓ and $t_5 = 15 – 8 = 7$ ✓

Now the question is one equation:

$$3n – 8 = 52 \ \Longrightarrow \ 3n = 60 \ \Longrightarrow \ n = 20$$

So yes, $52$ is the $20$th term.

This is exactly the trade between the two kinds of rule. The recursive form is the natural way to describe the sequence; the explicit form is the way to interrogate it.

Question 6

Let $T_1 = 1$, $T_2 = 2$, $T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \ge 4$. Find $T_4$, $T_5$, $T_6$, $T_7$ and $T_8$.

Solution. Each term is the sum of the three before it, so the calculation has to be built up one term at a time — there is no shortcut here, and the three starting values are all needed before the rule can fire at all.

$$T_4 = T_3 + T_2 + T_1 = 4 + 2 + 1 = 7$$

$$T_5 = T_4 + T_3 + T_2 = 7 + 4 + 2 = 13$$

$$T_6 = T_5 + T_4 + T_3 = 13 + 7 + 4 = 24$$

$$T_7 = T_6 + T_5 + T_4 = 24 + 13 + 7 = 44$$

$$T_8 = T_7 + T_6 + T_5 = 44 + 24 + 13 = 81$$

The sequence so far is

$$1, \ 2, \ 4, \ 7, \ 13, \ 24, \ 44, \ 81, \ \dots$$

This is the Tribonacci sequence — the three-term cousin of the Virahānka–Fibonacci sequence the chapter met earlier, where each term was the sum of the previous two. It is worth noticing that it grows faster: Virahānka’s sequence roughly multiplies by $1.618$ each step, this one by about $1.839$.

Unlike Question 5, there is no tidy explicit formula to fall back on here. Some recursive sequences convert; most do not.

Common mistakes

• Question 1, starting at $n = 0$. The rule says $n \ge 1$, so the first term is $t_1$. Using $n = 0$ shifts the whole list by one place.
• Question 1(iii), reading $n^2$ as $2n$. The terms $2, 3, 6, 11, 18$ come from squaring; $2n – 2n + 3$ would give the constant $3$ every time, which should be an immediate warning.
• Questions 3 and 4, forgetting to check that $n$ is a whole number. Solving $5n – 3 = 100$ gives $n = 20.6$, which means $100$ is not a term. The arithmetic alone is not the answer — the check is.
• Question 3, assuming one of the two must fail. Questions of this shape usually have one value in and one out. Here both are in, and the working says so.
• Question 5, climbing to $52$ one step at a time. It works, but it is twenty lines of addition with twenty chances to slip. Converting to $t_n = 3n – 8$ first is both faster and safer.
• Question 6, adding only the previous two terms. The rule names three. Adding two gives $1, 2, 4, 6, 10, 16, \dots$, which is a different sequence entirely.

Practise next

• Exercise Set 8.2 — arithmetic progressions proper, where the explicit rule $t_n = a + (n-1)d$ replaces the substitution done by hand here.
• The End-of-Chapter Exercises, Questions 14 and 15 — two more recursive sequences, one of which turns out to be an old friend in disguise.