NCERT Solutions Class 9 Maths Chapter 8 In-text Questions — Predicting What Comes Next: Exploring Sequences and Progressions

NCERT Class 9 Mathematics — Ganita Manjari Part I, Predicting What Comes Next: Exploring Sequences and Progressions. All 9 in-text Exercise prompts solved.

Also here. Solutions to the end-of-chapter questions for this chapter are on their own page.

Scattered through this chapter, between the worked examples, are nine short prompts marked Exercise: in bold. They are not part of any numbered Exercise Set, and they are easy to read past — but they are where each new idea is first put to work, and several are harder than the Exercise Set questions that follow them.

$$t_n = a + (n-1)d \qquad t_n = a\,r^{\,n-1} \qquad 1 + 2 + \dots + n = \frac{n(n+1)}{2}$$

Key insight. Before using any formula, check that the sequence is what you think it is. An AP must give the same difference for every consecutive pair, a GP the same ratio. Four of these nine prompts ask you to verify exactly that before writing an $n$th term, and the habit is what stops a formula being fitted to a sequence that does not have one.

Question 1

Consider the sequence $1, 4, 7, 10, 13, \dots$ 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?

Solution.

The next four terms. The differences are $3, 3, 3, 3$ — constant, so this is an AP with $a = 1$ and $d = 3$, and

$$t_n = 1 + (n-1)3 = 3n – 2$$

Continuing: $16, \ 19, \ 22, \ 25$.

The sequence of running totals. The new sequence has $S_n = t_1 + t_2 + \dots + t_n$, so each term is the previous total plus the next term of the original:

$n$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$t_n$	$1$	$4$	$7$	$10$	$13$	$16$	$19$	$22$	$25$	$28$
$S_n$	$1$	$5$	$12$	$22$	$35$	$51$	$70$	$92$	$117$	$145$

A closed form for the totals. Averaging first and last term, as with any AP:

$$S_n = \frac{n}{2}\big(1 + (3n-2)\big) = \frac{n(3n-1)}{2}$$

Check at $n = 10$: $\frac{10 \times 29}{2} = 145$ ✓

Notice that $S_n$ is quadratic in $n$ where $t_n$ was linear. Summing an AP always raises the degree by one, which is the same thing that makes the triangular numbers quadratic. (These particular totals $1, 5, 12, 22, 35, \dots$ are the pentagonal numbers.)

Question 2

Can you write $t_5$, $t_6$, $t_7$ and $t_8$ for the sequence of triangular numbers?

Solution. The triangular numbers are $1, 3, 6, 10, 15, \dots$, the $n$th being the sum of the first $n$ natural numbers:

$$t_n = \frac{n(n+1)}{2}$$

$$t_5 = \frac{5 \times 6}{2} = 15, \qquad t_6 = \frac{6 \times 7}{2} = 21$$

$$t_7 = \frac{7 \times 8}{2} = 28, \qquad t_8 = \frac{8 \times 9}{2} = 36$$

Check them against the pattern rather than only against the formula: the differences should be $6, 7, 8$ — since each triangular number adds the next natural number — and $21 – 15 = 6$, $28 – 21 = 7$, $36 – 28 = 8$ ✓

Question 3

Using the explicit rule $u_n = 2n – 1$, find the $53$rd term, the $108$th term, and the $1170$th term of the odd number sequence.

Solution. This is exactly what an explicit rule buys you — no need to generate anything in between:

$$u_{53} = 2(53) – 1 = 106 – 1 = 105$$

$$u_{108} = 2(108) – 1 = 216 – 1 = 215$$

$$u_{1170} = 2(1170) – 1 = 2340 – 1 = 2339$$

A recursive rule would have needed $1169$ additions to reach the last of these.

Question 4

Consider the expression $t_n = 3n – 7$.

Solution.

(i) Find its first, second, third, $12$th, $18$th and $50$th terms

$$t_1 = 3 – 7 = -4, \qquad t_2 = 6 – 7 = -1, \qquad t_3 = 9 – 7 = 2$$

$$t_{12} = 36 – 7 = 29, \qquad t_{18} = 54 – 7 = 47, \qquad t_{50} = 150 – 7 = 143$$

The rule is linear in $n$, so this is an AP with $d = 3$ — and indeed the first three terms rise by $3$ each time, starting from a negative value.

(ii) Which term of the sequence is $332$?

$$3n – 7 = 332 \ \Longrightarrow \ 3n = 339 \ \Longrightarrow \ n = 113$$

$n = 113$ is a positive whole number, so $332$ is the $113$th term. Check: $3(113) – 7 = 339 – 7 = 332$ ✓

(iii) Is $557$ a term of this sequence? Why or why not?

$$3n – 7 = 557 \ \Longrightarrow \ 3n = 564 \ \Longrightarrow \ n = 188$$

$188$ is a positive whole number, so yes — $557$ is the $188$th term. Check: $3(188) – 7 = 564 – 7 = 557$ ✓

Do not be steered by the phrasing. “Why or why not?” reads as an invitation to answer no, and questions of this shape usually are — but here the arithmetic simply works out. The same thing happens in Exercise Set 8.1 Question 3, where both offered values turn out to be terms. The test is the arithmetic, never the wording.

What a “no” would look like. Take $558$ instead: $3n = 565$, and $565 \div 3$ is not a whole number, so $558$ is not a term. Equivalently, every term of this sequence leaves remainder $2$ on division by $3$ (since $3n – 7 = 3(n-3) + 2$), and $557 = 3 \times 185 + 2$ ✓ while $558 = 3 \times 186 + 0$ ✗.

Question 5

Verify that the following sequences are arithmetic progressions and write their $n$th terms. What do you observe when you plot the ordered pairs emerging from them?

Solution.

(i) $2, 5, 8, 11, \dots$

Differences: $5 – 2 = 3$, $8 – 5 = 3$, $11 – 8 = 3$. All equal, so it is an AP with $a = 2$, $d = 3$:

$$t_n = 2 + (n-1)3 = 3n – 1$$

(ii) $-5, -1, 3, 7, \dots$

Differences: $-1 – (-5) = 4$, $3 – (-1) = 4$, $7 – 3 = 4$. All equal, so it is an AP with $a = -5$, $d = 4$:

$$t_n = -5 + (n-1)4 = 4n – 9$$

Check: $t_1 = -5$ ✓ and $t_4 = 16 – 9 = 7$ ✓

What you observe on plotting

Plot the ordered pairs $(n, t_n)$ — $(1,2), (2,5), (3,8), (4,11)$ for the first, and $(1,-5), (2,-1), (3,3), (4,7)$ for the second. In both cases the points lie on a straight line.

That is not a coincidence, and it is the real content of the question. An AP’s $n$th term is $t_n = dn + (a – d)$ — a linear expression in $n$ — so plotting $t_n$ against $n$ is plotting a straight line, with the common difference $d$ as its slope. The second sequence climbs more steeply than the first because $4 > 3$.

The converse is the useful half: if the plotted points of a sequence do not lie on a line, the sequence is not an AP, and no amount of searching will produce a common difference.

Question 6

Using the formula $t_n = a + (n-1) \times d$, find the $n$th term of the following arithmetic progressions.

Solution.

(i) $\dfrac{1}{2}, \ \dfrac{5}{2}, \ \dfrac{9}{2}, \ \dfrac{13}{2}, \dots$

Every term has denominator $2$, so compare numerators: $1, 5, 9, 13$ rise by $4$, meaning the terms rise by $\frac42 = 2$.

$$a = \frac12, \qquad d = \frac{5}{2} – \frac{1}{2} = 2$$

$$t_n = \frac12 + (n-1)2 = 2n – 2 + \frac12 = 2n – \frac32 = \frac{4n-3}{2}$$

Check: $t_1 = \frac12$ ✓, $t_4 = \frac{16-3}{2} = \frac{13}{2}$ ✓

(ii) $1.5, \ 3.5, \ 5.5, \ 7.5, \dots$

$$a = 1.5, \qquad d = 3.5 – 1.5 = 2$$

$$t_n = 1.5 + (n-1)2 = 2n – 0.5 = \frac{4n-1}{2}$$

Check: $t_1 = 1.5$ ✓, $t_4 = 7.5$ ✓

The two are the same AP shifted. Both have $d = 2$; the second is the first plus $1$ throughout. A fractional or decimal first term changes nothing about the method — the formula never assumed whole numbers.

Question 7

Find recursive rules for the APs in the previous exercises.

Solution. A recursive rule for an AP is always the same two lines — name the first term, then add the common difference — so the work is already done once $a$ and $d$ are known:

$$t_1 = a, \qquad t_{n+1} = t_n + d$$

For the four APs of Questions 5 and 6:

AP	$a$	$d$	Recursive rule
$2, 5, 8, 11, \dots$	$2$	$3$	$t_1 = 2$, $\ t_{n+1} = t_n + 3$
$-5, -1, 3, 7, \dots$	$-5$	$4$	$t_1 = -5$, $\ t_{n+1} = t_n + 4$
$\frac12, \frac52, \frac92, \frac{13}{2}, \dots$	$\frac12$	$2$	$t_1 = \frac12$, $\ t_{n+1} = t_n + 2$
$1.5, 3.5, 5.5, 7.5, \dots$	$1.5$	$2$	$t_1 = 1.5$, $\ t_{n+1} = t_n + 2$

The first term is not optional. “$t_{n+1} = t_n + 2$” on its own describes every AP with common difference $2$ — the third and fourth rows above have identical step rules and are different sequences. Only $t_1$ separates them.

Question 8

Check whether the following sequences are geometric progressions and find their $n$th terms.

Solution. The test for a GP is that every consecutive pair gives the same ratio — so divide each term by the one before it, right across the printed terms.

(i) $2, 10, 50, 250, \dots$

$$\frac{10}{2} = 5, \qquad \frac{50}{10} = 5, \qquad \frac{250}{50} = 5$$

A GP with $a = 2$, $r = 5$:

$$t_n = 2 \times 5^{\,n-1}$$

(ii) $4, \ \dfrac{8}{3}, \ \dfrac{16}{9}, \ \dfrac{32}{27}, \dots$

$$\frac{8/3}{4} = \frac{8}{12} = \frac23, \qquad \frac{16/9}{8/3} = \frac{16}{9} \times \frac{3}{8} = \frac23, \qquad \frac{32/27}{16/9} = \frac{32}{27} \times \frac{9}{16} = \frac23$$

A GP with $a = 4$, $r = \frac23$:

$$t_n = 4\left(\frac23\right)^{\,n-1}$$

Because $r < 1$ the terms decrease towards $0$ without ever reaching it — a GP need not grow.

(iii) $3, \ -\dfrac{3}{2}, \ \dfrac{3}{4}, \ -\dfrac{3}{8}, \dots$

$$\frac{-3/2}{3} = -\frac12, \qquad \frac{3/4}{-3/2} = -\frac12, \qquad \frac{-3/8}{3/4} = -\frac12$$

A GP with $a = 3$, $r = -\frac12$:

$$t_n = 3\left(-\frac12\right)^{\,n-1}$$

The negative ratio is what makes the signs alternate, and it is perfectly legitimate: nothing in the definition of a GP requires $r > 0$. The terms shrink in size while flipping sign each step.

Question 9

Can you find a recursive rule for the formula $t_n = 3 \times 10^{\,n-1}$ that generates the geometric progression $3, 30, 300, 3000, \dots$?

Solution. Read the ratio straight off the explicit formula: raising $n$ by $1$ multiplies $10^{\,n-1}$ by $10$, so each term is ten times the one before.

$$t_1 = 3, \qquad t_{n+1} = 10\,t_n \quad \text{for } n \ge 1$$

Verify it reproduces the sequence: $t_1 = 3$, then $30$, $300$, $3000$ ✓

In general, an explicit GP rule $t_n = a r^{\,n-1}$ converts to $t_1 = a$, $t_{n+1} = r\,t_n$, and the conversion runs both ways. Here is the algebra, if you want it rather than the pattern:

$$\frac{t_{n+1}}{t_n} = \frac{3 \times 10^{\,n}}{3 \times 10^{\,n-1}} = 10$$

Common mistakes

• Question 1, giving only the next four terms. The prompt asks for two sequences — the continuation and the ten running totals. The second half is the part that matters.
• Question 3, computing $2 \times 53$ and stopping. $106$ is not the answer; the rule subtracts $1$. Always finish the substitution.
• Question 4(iii), answering “no” because the question hints at it. $3n – 7 = 557$ gives the whole number $n = 188$, so $557$ is a term. Do the division before deciding.
• Question 5, checking one difference only. A sequence whose first two gaps match may still fail at the third. “Verify” means check every printed pair.
• Question 5, missing the point of the plot. The observation asked for is that the points are collinear — that is what “arithmetic” looks like on a graph.
• Question 6, being put off by the fractions. $\frac52 – \frac12 = 2$ is an ordinary common difference. The formula never required whole numbers.
• Question 7, giving the step rule without the first term. $t_{n+1} = t_n + 2$ describes infinitely many sequences. A recursive rule is incomplete without $t_1$.
• Question 8, testing differences instead of ratios. $4, \frac83, \frac{16}{9}$ has no common difference and is still a perfectly good GP.
• Question 8(iii), rejecting it because the terms alternate in sign. A negative common ratio is allowed, and it is exactly what alternating signs mean.

Practise next

• Exercise Set 8.1 — the same explicit-and-recursive machinery as Questions 3, 4, 7 and 9, set as a numbered exercise.
• Exercise Set 8.2 and Exercise Set 8.3 — APs and GPs in full, following Questions 5 to 8.
• The Think and Reflect prompts for this chapter — the conceptual companions to these, on why an explicit formula is worth having at all.