NCERT Solutions Class 9 Maths Chapter 8 End-of-Chapter Exercises — Predicting What Comes Next: Exploring Sequences and Progressions

NCERT Class 9 Mathematics — Ganita Manjari Part I, Predicting What Comes Next: Exploring Sequences and Progressions, End-of-Chapter Exercises. All 15 questions solved.

APs and GPs are mixed together here without warning, so the first job on every question is to decide which you are looking at. Four formulae cover the lot:

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

Every question in this set carries an asterisk in the book, which marks them as not assessed.

Key insight. When two facts about a progression are given, write each as an equation in $a$ and $d$ (or $a$ and $r$) and eliminate. For an AP, subtracting kills $a$; for a GP, dividing kills $a$. Questions 1, 2, 5, 8, 11 and 13 are all that one move, followed by tidying up.

Question 1

Find the $31$st term of an AP whose $11$th term is $38$ and $16$th term is $73$.

Solution.

$$t_{11} = a + 10d = 38, \qquad t_{16} = a + 15d = 73$$

Subtracting eliminates $a$:

$$5d = 35 \ \Longrightarrow \ d = 7$$

$$a = 38 – 10(7) = 38 – 70 = -32$$

$$t_{31} = a + 30d = -32 + 210 = 178$$

There is a shorter route worth knowing: from the $16$th term to the $31$st is $15$ steps, the same gap as from the $11$th to the $16$th multiplied by three — so $t_{31} = 73 + 15(7) = 178$ ✓

Question 2

Determine the AP whose third term is $16$ and whose $7$th term exceeds the $5$th term by $12$.

Solution. The second fact is a statement about the gap between two terms, and in an AP that gap depends only on $d$:

$$t_7 – t_5 = (a + 6d) – (a + 4d) = 2d = 12 \ \Longrightarrow \ d = 6$$

Then from $t_3 = a + 2d = 16$:

$$a = 16 – 12 = 4$$

So the AP is

$$4, \ 10, \ 16, \ 22, \ 28, \ \dots$$

Check both conditions: $t_3 = 16$ ✓, and $t_7 – t_5 = 40 – 28 = 12$ ✓

Question 3

How many three-digit numbers are divisible by $7$?

Solution. Three-digit numbers run from $100$ to $999$, and those divisible by $7$ form an AP of common difference $7$.

Smallest. $100 \div 7 = 14$ remainder $2$, so $98 = 7 \times 14$ is too small and the first three-digit multiple is $7 \times 15 = 105$.

Largest. $999 \div 7 = 142$ remainder $5$, so the last is $7 \times 142 = 994$.

How many. Solve $105 + (n-1)7 = 994$:

$$(n-1)7 = 889 \ \Longrightarrow \ n – 1 = 127 \ \Longrightarrow \ n = 128$$

Equivalently, count the multiplier: the multiples are $7 \times 15$ up to $7 \times 142$, and $142 – 15 + 1 = 128$ ✓

Question 4

How many multiples of $4$ lie between $10$ and $250$?

Solution. The multiples of $4$ in this range form an AP with $d = 4$.

Smallest multiple of $4$ above $10$: $12$. Largest below $250$: $248$ (since $250 \div 4 = 62.5$, and $4 \times 62 = 248$).

$$12 + (n-1)4 = 248 \ \Longrightarrow \ (n-1)4 = 236 \ \Longrightarrow \ n = 60$$

Or by multiplier: $4 \times 3$ up to $4 \times 62$, giving $62 – 3 + 1 = 60$ ✓

Question 5

Find a GP for which the sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.

Solution. Write both facts with $t_n = a r^{\,n-1}$.

The second fact first, because it involves only $r$:

$$t_5 = 4\,t_3 \ \Longrightarrow \ a r^4 = 4 a r^2$$

Dividing by $ar^2$ (neither $a$ nor $r$ can be $0$ in a genuine GP):

$$r^2 = 4 \ \Longrightarrow \ r = 2 \ \text{ or } \ r = -2$$

Now the first fact, $a + ar = -4$, taken in each case.

Case $r = 2$: $a(1 + 2) = -4$, so $a = -\frac43$ and the GP is

$$-\frac43, \ -\frac83, \ -\frac{16}{3}, \ -\frac{32}{3}, \ \dots$$

Check: sum of first two $= -\frac43 – \frac83 = -4$ ✓; $t_3 = -\frac{16}{3}$ and $t_5 = -\frac{64}{3} = 4 t_3$ ✓

Case $r = -2$: $a(1 – 2) = -4$, so $-a = -4$, $a = 4$ and the GP is

$$4, \ -8, \ 16, \ -32, \ 64, \ \dots$$

Check: $4 + (-8) = -4$ ✓; $t_3 = 16$ and $t_5 = 64 = 4 \times 16$ ✓

Both are valid answers. The second is the tidier one and is what most solutions quote, but nothing in the question rules out the first. (Solving the equations blindly also produces $r = 0$ with $a = -4$; that is not a GP — a common ratio of $0$ collapses the sequence to $-4, 0, 0, 0, \dots$ — so it is discarded.)

Question 6

Find all possible ways of expressing $100$ as the sum of consecutive natural numbers.

Solution. Suppose the sum runs from $a$ for $k$ terms, where $k \ge 2$. It is an AP with $d = 1$, so

$$S = \frac{k}{2}\,\big(2a + k – 1\big) = 100 \ \Longrightarrow \ k\,(2a + k – 1) = 200$$

Write $m = 2a + k – 1$, so $km = 200$ and

$$a = \frac{m – k + 1}{2}$$

The key observation. $k$ and $m$ have opposite parity: if $k$ is even then $k – 1$ is odd, making $m = 2a + k – 1$ odd, and vice versa. Since $200 = 2^3 \times 5^2$, its only odd factors are $1, 5, 25$ — so one of $k, m$ must be one of those three, and that leaves very few possibilities to test.

$k$	$m = 200/k$	$a = \frac{m-k+1}{2}$	Valid?
$5$	$40$	$18$	✓
$25$	$8$	$-8$	✗ ($a$ not a natural number)
$8$	$25$	$9$	✓
$40$	$5$	$-17$	✗
$200$	$1$	$-99$	✗

So there are exactly two ways:

$$18 + 19 + 20 + 21 + 22 = 100 \qquad (5 \text{ terms, average } 20)$$

$$9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100 \qquad (8 \text{ terms, average } 12.5)$$

Why there are so few. Every odd factor $k$ of $200$ gives a candidate, and $200$ has only three odd factors because it is $8 \times 25$. This is a genuine theorem in miniature: a number can be written as a sum of consecutive naturals in as many ways as it has odd factors greater than $1$ (subject to the terms staying positive), which is why powers of $2$ cannot be written that way at all.

Question 7

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 $2$nd hour, $4$th hour and $n$th hour?

Solution. Doubling every hour is multiplication by a fixed factor, so the counts form a GP with $r = 2$ starting from $30$ at hour $0$:

$$N_n = 30 \times 2^{\,n}$$

where $n$ counts hours elapsed, so the exponent is the hour number itself.

$$N_2 = 30 \times 2^2 = 30 \times 4 = 120$$

$$N_4 = 30 \times 2^4 = 30 \times 16 = 480$$

Check by stepping: $30 \to 60 \to 120 \to 240 \to 480$ — hours $0, 1, 2, 3, 4$ ✓

Watch the index. If you set up the GP with $a = 30$ as the “first term” and use $t_n = a r^{\,n-1}$, then $t_n$ is the count at the end of hour $n – 1$, and the answer for the $2$nd hour would be $t_3$. Both conventions work; mixing them gives $60$ and $240$, which are the answers to the neighbouring questions.

Question 8

The sum of the $4$th and $8$th terms of an AP is $24$ and the sum of the $6$th and $10$th terms is $44$. Find the first three terms of the AP.

Solution.

$$t_4 + t_8 = (a + 3d) + (a + 7d) = 2a + 10d = 24 \ \Longrightarrow \ a + 5d = 12$$

$$t_6 + t_{10} = (a + 5d) + (a + 9d) = 2a + 14d = 44 \ \Longrightarrow \ a + 7d = 22$$

Subtracting:

$$2d = 10 \ \Longrightarrow \ d = 5, \qquad a = 12 – 25 = -13$$

So the first three terms are

$$-13, \ -8, \ -3$$

Check: $t_4 = 2$, $t_8 = 22$, sum $24$ ✓; $t_6 = 12$, $t_{10} = 32$, sum $44$ ✓

Notice the shortcut hiding in the first line: $t_4 + t_8 = 2(a + 5d) = 2\,t_6$. So $t_6 = 12$ and $t_8 = 22$ come out directly, and $d = \frac{22 – 12}{2} = 5$ without solving anything.

Question 9

Find the smallest value of $n$ such that the sum of the first $n$ natural numbers is greater than $1{,}000$.

Solution. The sum of the first $n$ natural numbers is

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

so the condition is

$$\frac{n(n+1)}{2} > 1000 \ \Longrightarrow \ n(n+1) > 2000$$

Since $n(n+1)$ is close to $n^2$, try $n \approx \sqrt{2000} \approx 44.7$:

$$n = 44: \ 44 \times 45 = 1980 \ \ (\text{not more than } 2000)$$ $$n = 45: \ 45 \times 46 = 2070 \ \ (\text{more than } 2000) \ \checkmark$$

So the smallest such $n$ is $45$. The sums themselves are $S_{44} = 990$ and $S_{45} = 1035$ — the crossing happens between them.

Question 10

Which term of the GP $2, 8, 32, \dots$ is $131072$? Write the explicit formula as well as the recursive formula for the $n^{th}$ term.

Solution.

$$a = 2, \qquad r = \frac{8}{2} = 4$$

Explicit: $t_n = 2 \times 4^{\,n-1}$.   Recursive: $t_1 = 2$, $t_{n+1} = 4t_n$.

Which term is $131072$?

$$2 \times 4^{\,n-1} = 131072 \ \Longrightarrow \ 4^{\,n-1} = 65536$$

Now write $65536$ as a power of $4$. Since $65536 = 2^{16}$ and $4 = 2^2$:

$$4^{\,n-1} = 2^{16} = 4^{8} \ \Longrightarrow \ n – 1 = 8 \ \Longrightarrow \ n = 9$$

Check: $2 \times 4^8 = 2 \times 65536 = 131072$ ✓

Question 11

The sum of the first three terms of a GP is $\dfrac{13}{12}$ and their product is $-1$. Find the common ratio and the terms.

Solution. When three terms of a GP are given by their product, write them symmetrically as

$$\frac{a}{r}, \quad a, \quad ar$$

because then the product is simply $a^3$ — the $r$’s cancel. This is the standard trick and it is what makes the question tractable.

From the product:

$$\frac{a}{r} \times a \times ar = a^3 = -1 \ \Longrightarrow \ a = -1$$

From the sum:

$$\frac{-1}{r} – 1 – r = \frac{13}{12} \ \Longrightarrow \ -\left(\frac1r + r\right) = \frac{13}{12} + 1 = \frac{25}{12}$$

$$\frac1r + r = -\frac{25}{12} \ \Longrightarrow \ 12 + 12r^2 = -25r \ \Longrightarrow \ 12r^2 + 25r + 12 = 0$$

Factorising (or by the formula, with discriminant $625 – 576 = 49$):

$$r = \frac{-25 \pm 7}{24} \ \Longrightarrow \ r = -\frac34 \ \text{ or } \ r = -\frac43$$

The terms. Taking $r = -\frac34$ with $a = -1$:

$$\frac{a}{r} = \frac{-1}{-3/4} = \frac43, \qquad a = -1, \qquad ar = \frac34$$

so the GP is $\ \frac43, \ -1, \ \frac34$.

Taking $r = -\frac43$ gives $\ \frac34, \ -1, \ \frac43$ — the same three numbers in the opposite order, which is why the quadratic has two roots that are reciprocals of each other.

Check: $\frac43 – 1 + \frac34 = \frac{16 – 12 + 9}{12} = \frac{13}{12}$ ✓ and $\frac43 \times (-1) \times \frac34 = -1$ ✓

Question 12

If the $4$th, $10$th and $16$th terms of a GP are $x$, $y$ and $z$ respectively, prove that $x$, $y$, $z$ are in GP.

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

$$x = a r^{3}, \qquad y = a r^{9}, \qquad z = a r^{15}$$

To show three numbers are in GP, show that consecutive ratios are equal:

$$\frac{y}{x} = \frac{a r^{9}}{a r^{3}} = r^{6}, \qquad \frac{z}{y} = \frac{a r^{15}}{a r^{9}} = r^{6}$$

The two ratios are equal, so $x$, $y$, $z$ form a GP with common ratio $r^{6}$. $\blacksquare$

Equivalently, in the $y^2 = xz$ form:

$$y^2 = \left(a r^{9}\right)^2 = a^2 r^{18}, \qquad xz = a r^{3} \cdot a r^{15} = a^2 r^{18}$$

so $y^2 = xz$, which is the condition for three numbers to be in GP.

The general fact behind it. The positions $4, 10, 16$ are themselves in arithmetic progression, with common difference $6$. Whenever you pick terms of a GP at equally spaced positions, the terms you pick are again a GP — and the new ratio is the old one raised to the power of that spacing. Nothing about $4$, $10$ and $16$ in particular was used.

Question 13

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.

Solution. Take the terms as $a$, $ar$, $ar^2$. The two conditions are

$$a\left(1 + r + r^2\right) = 26 \qquad\text{(i)}$$ $$a^2\left(1 + r^2 + r^4\right) = 364 \qquad\text{(ii)}$$

The move that makes this easy is the factorisation

$$1 + r^2 + r^4 = \left(1 + r + r^2\right)\left(1 – r + r^2\right)$$

(Multiply the brackets out to check — the $r$ and $r^3$ terms cancel.) Substituting into (ii):

$$a^2\left(1 + r + r^2\right)\left(1 – r + r^2\right) = 364$$

Now divide by the square of (i), namely $a^2(1+r+r^2)^2 = 676$:

$$\frac{1 – r + r^2}{1 + r + r^2} = \frac{364}{676} = \frac{7}{13}$$

Cross-multiplying:

$$13 – 13r + 13r^2 = 7 + 7r + 7r^2 \ \Longrightarrow \ 6r^2 – 20r + 6 = 0 \ \Longrightarrow \ 3r^2 – 10r + 3 = 0$$

$$(3r – 1)(r – 3) = 0 \ \Longrightarrow \ r = 3 \ \text{ or } \ r = \frac13$$

Case $r = 3$: from (i), $a(1 + 3 + 9) = 26$, so $13a = 26$ and $a = 2$. The terms are $\ 2, \ 6, \ 18$.

Case $r = \frac13$: the same three numbers in reverse, $\ 18, \ 6, \ 2$.

Check: $2 + 6 + 18 = 26$ ✓ and $4 + 36 + 324 = 364$ ✓

Question 14

Suppose $P_1 = 1$, $P_2 = 2$ and for $n > 2$, $P_n = P_1 + P_2 + \dots + P_{n-1} + 1$. Find the values of $P_1, P_2, \dots, P_8$. Can you find a simpler recursive formula for $P_n$? Can you give an explicit formula?

Solution. Each term is one more than the sum of everything before it.

$$P_3 = (1 + 2) + 1 = 4$$ $$P_4 = (1 + 2 + 4) + 1 = 8$$ $$P_5 = (1 + 2 + 4 + 8) + 1 = 16$$ $$P_6 = 31 + 1 = 32, \qquad P_7 = 63 + 1 = 64, \qquad P_8 = 127 + 1 = 128$$

$$1, \ 2, \ 4, \ 8, \ 16, \ 32, \ 64, \ 128$$

A simpler recursive formula. The list doubles, so

$$P_{n+1} = 2\,P_n \quad \text{with } P_1 = 1$$

Why doubling happens. Write $S_{n-1} = P_1 + \dots + P_{n-1}$. Then $P_n = S_{n-1} + 1$ and $P_{n+1} = S_n + 1 = S_{n-1} + P_n + 1$. Subtracting the first from the second:

$$P_{n+1} – P_n = P_n \ \Longrightarrow \ P_{n+1} = 2P_n$$

So the doubling is not a guess from the first few terms — it follows from the rule itself.

An explicit formula. A GP with first term $1$ and ratio $2$:

$$P_n = 2^{\,n-1}$$

Question 15

Suppose $W_1 = 1$, $W_2 = 2$ and for $n > 2$, $W_n = W_1 + W_2 + \dots + W_{n-2} + 2$. Find the values of $W_1, W_2, \dots, W_8$. Do you recognise this sequence?

Solution. Read the rule carefully. The sum stops at $W_{n-2}$, not $W_{n-1}$ — the term immediately before $W_n$ is left out. That single change is what makes this sequence completely different from Question 14’s.

$$W_3 = W_1 + 2 = 1 + 2 = 3$$ $$W_4 = W_1 + W_2 + 2 = 1 + 2 + 2 = 5$$ $$W_5 = W_1 + W_2 + W_3 + 2 = 1 + 2 + 3 + 2 = 8$$ $$W_6 = 1 + 2 + 3 + 5 + 2 = 13$$ $$W_7 = 1 + 2 + 3 + 5 + 8 + 2 = 21$$ $$W_8 = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34$$

$$1, \ 2, \ 3, \ 5, \ 8, \ 13, \ 21, \ 34$$

Do you recognise it? Yes — this is the Virahānka–Fibonacci sequence, met earlier in the chapter, in which each term is the sum of the previous two.

Why the two rules agree. Let $S_{n-2} = W_1 + \dots + W_{n-2}$, so $W_n = S_{n-2} + 2$ and $W_{n+1} = S_{n-1} + 2 = S_{n-2} + W_{n-1} + 2$. Subtracting:

$$W_{n+1} – W_n = W_{n-1} \ \Longrightarrow \ W_{n+1} = W_n + W_{n-1}$$

which is exactly Virahānka’s rule. So the strange-looking definition is the familiar one wearing a disguise — and Questions 14 and 15 together make the point that a tiny change to a recursive rule (stopping the sum one term earlier) can change doubling into Fibonacci growth.

Common mistakes

• Question 3, taking $100$ or $1000$ as an endpoint. The first three-digit multiple of $7$ is $105$ and the last is $994$; neither boundary is itself a multiple.
• Question 4, including $250$ or $10$. “Between” excludes both, and neither is a multiple of $4$ anyway — but $248$ and $12$ must be found, not guessed.
• Question 5, stopping at $r = 2$. $r^2 = 4$ has two roots, and both give a valid GP. The $r = 0$ root that also drops out of the algebra is not a GP and must be discarded.
• Question 6, counting $100$ itself or reversed runs. “Consecutive natural numbers” means at least two of them, increasing; $22 + 21 + 20 + 19 + 18$ is the same sum, not a second way.
• Question 7, using $2^{\,n-1}$. With $30$ bacteria at hour $0$, the count at hour $n$ is $30 \times 2^{\,n}$. The two conventions differ by a factor of $2$ throughout.
• Question 11, taking the terms as $a$, $ar$, $ar^2$. It works but the product becomes $a^3r^3$, which is messier. The symmetric form $\frac{a}{r}, a, ar$ makes the product $a^3$ and the question falls out in two lines.
• Question 13, squaring the sum. $(a + ar + ar^2)^2$ is not the sum of the squares — it carries cross terms. The factorisation of $1 + r^2 + r^4$ is the route.
• Question 15, summing to $W_{n-1}$. That is Question 14’s rule and gives $1, 2, 4, 8, \dots$ again. The stop at $W_{n-2}$ is the entire point of the question.

Practise next

• Exercise Set 8.2 — the AP formulae on their own, if Questions 1, 2 and 8 were the awkward ones.
• Exercise Set 8.3 — GPs on their own, including the Sierpiński carpet, which is Question 14’s doubling idea in a geometric setting.
• Exercise Set 8.1, Question 6 — the Tribonacci rule, a natural companion to Question 15 here.