NCERT Solutions Class 9 Maths Exercise 8.2 — 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 8.2. All 7 questions solved.

An arithmetic progression is a sequence in which each term exceeds the one before by a fixed amount — the common difference $d$. Two formulae run this entire exercise.

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

where $a$ is the first term, $\ell$ the last, and $n$ the number of terms. The second is Āryabhaṭa’s rule quoted in the chapter: the average of the first and last terms, multiplied by how many there are.

Key insight. Almost every AP question reduces to finding $a$ and $d$, because everything else follows from them. Given any two terms you can always get both: subtracting $t_p = a + (p-1)d$ from $t_q = a + (q-1)d$ kills $a$ and leaves $t_q – t_p = (q-p)d$, so the common difference is the gap in values divided by the gap in positions.

Question 1

Find the $10^{th}$ and $26^{th}$ terms of the AP: $3, 8, 13, 18, \dots$

Solution. Read off the two quantities that matter:

$$a = 3, \qquad d = 8 – 3 = 5$$

(Confirm $d$ with another pair: $13 – 8 = 5$ ✓ and $18 – 13 = 5$ ✓ — an AP must give the same difference everywhere, and checking one gap only is how a non-AP slips through.)

$$t_{10} = a + 9d = 3 + 9(5) = 3 + 45 = 48$$

$$t_{26} = a + 25d = 3 + 25(5) = 3 + 125 = 128$$

Note the $(n-1)$, not $n$: reaching the $10$th term takes nine steps from the first.

Question 2

Which term of the AP $21, 18, 15, \dots$ is $-81$? Also, is $0$ a term of this AP? Give reasons for your answer.

Solution.

$$a = 21, \qquad d = 18 – 21 = -3$$

so the AP is decreasing, and

$$t_n = 21 + (n-1)(-3) = 21 – 3n + 3 = 24 – 3n$$

Which term is $-81$?

$$24 – 3n = -81 \ \Longrightarrow \ 3n = 105 \ \Longrightarrow \ n = 35$$

So $-81$ is the $35$th term. Check: $t_{35} = 24 – 105 = -81$ ✓

Is $0$ a term?

$$24 – 3n = 0 \ \Longrightarrow \ n = 8$$

$n = 8$ is a positive whole number, so yes — $0$ is the $8$th term. Writing out the AP makes it plain:

$$21, \ 18, \ 15, \ 12, \ 9, \ 6, \ 3, \ \mathbf{0}, \ -3, \ \dots$$

The reason it works. Every term of this AP is a multiple of $3$ (since $24 – 3n = 3(8 – n)$), and $0$ is a multiple of $3$. Had the question asked about $1$ or $-8$, the equation would have given a fractional $n$ and the answer would have been no.

Question 3

Find the $n^{th}$ term of the AP $11, 8, 5, 2 \dots$ Write the recursive rule for this AP.

Solution.

$$a = 11, \qquad d = 8 – 11 = -3$$

The explicit rule.

$$t_n = a + (n-1)d = 11 + (n-1)(-3) = 11 – 3n + 3 = 14 – 3n$$

Check against the given terms: $t_1 = 14 – 3 = 11$ ✓, $t_2 = 14 – 6 = 8$ ✓, $t_4 = 14 – 12 = 2$ ✓

The recursive rule. A recursive rule needs a starting value and a step:

$$t_1 = 11, \qquad t_{n+1} = t_n – 3 \quad \text{for } n \ge 1$$

(Equivalently $t_n = t_{n-1} – 3$ for $n \ge 2$ — the same rule with the index shifted.)

The two descriptions carry exactly the same information. The explicit one answers “what is $t_{50}$?” in a line; the recursive one says more directly what the sequence does, which is subtract $3$ each time.

Question 4

An AP consists of $50$ terms in which the $3$rd term is $12$ and the last term is $106$. Find the $29$th term.

Solution. “The last term” of a $50$-term AP means $t_{50}$. Writing both facts with the formula gives the pair of equations the hint names:

$$t_3 = a + 2d = 12$$ $$t_{50} = a + 49d = 106$$

Subtract the first from the second — this is the standard move, because it eliminates $a$:

$$47d = 94 \ \Longrightarrow \ d = 2$$

Then from $a + 2d = 12$:

$$a = 12 – 4 = 8$$

Now the term asked for:

$$t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64$$

Check both given facts against $a = 8$, $d = 2$: $t_3 = 8 + 4 = 12$ ✓ and $t_{50} = 8 + 98 = 106$ ✓

Question 5

How many $2$-digit numbers are divisible by $3$? What is the sum of all these $2$-digit numbers?

Solution. The two-digit numbers run from $10$ to $99$. Among them, the multiples of $3$ form an AP.

The first and last. The smallest two-digit multiple of $3$ is $12$ ($10$ and $11$ are not multiples); the largest is $99$. So

$$a = 12, \qquad d = 3, \qquad \ell = 99$$

How many? Solve $t_n = \ell$:

$$12 + (n-1)3 = 99 \ \Longrightarrow \ (n-1)3 = 87 \ \Longrightarrow \ n – 1 = 29 \ \Longrightarrow \ n = 30$$

So there are $30$ of them.

Their sum. By the average-times-count rule:

$$S_{30} = \frac{30}{2}\,(12 + 99) = 15 \times 111 = 1665$$

A cross-check from a different direction: all multiples of $3$ up to $99$ sum to $3(1 + 2 + \dots + 33) = 3 \times \frac{33 \times 34}{2} = 3 \times 561 = 1683$, and the one-digit ones ($3, 6, 9$) contribute $18$. So the two-digit ones give $1683 – 18 = 1665$ ✓

Question 6

Harish started work at an annual salary of ₹$5{,}00{,}000$ and received an increment of ₹$20{,}000$ each year. After how many years did his income reach ₹$7{,}00{,}000$?

Solution. His annual salaries form an AP:

$$a = 5{,}00{,}000, \qquad d = 20{,}000$$

$$t_n = 5{,}00{,}000 + (n-1)(20{,}000)$$

Set this equal to the target:

$$5{,}00{,}000 + 20{,}000(n-1) = 7{,}00{,}000$$

$$20{,}000(n-1) = 2{,}00{,}000 \ \Longrightarrow \ n – 1 = 10 \ \Longrightarrow \ n = 11$$

Reading the answer correctly. $n = 11$ means ₹$7{,}00{,}000$ is his salary in the $11$th year of work. The question asks after how many years, and the eleventh year’s salary arrives after $10$ years of increments — one increment at the end of each of years $1$ to $10$.

The arithmetic makes this obvious in a way the formula does not: he needs to gain ₹$2{,}00{,}000$ at ₹$20{,}000$ a year, and $2{,}00{,}000 \div 20{,}000 = 10$.

Question 7

A child arranges marbles in rows so that the first row has $1$ marble, the second has $2$ marbles, the third has $3$, and so on up to $25$ rows. How many marbles does the child use in all?

Solution. The row counts are $1, 2, 3, \dots, 25$ — an AP with $a = 1$, $d = 1$, $\ell = 25$ and $n = 25$ terms. The total is their sum:

$$S_{25} = \frac{25}{2}\,(1 + 25) = \frac{25 \times 26}{2} = 25 \times 13 = 325$$

This is the $25$th triangular number, and the picture explains the name: the marbles really do form a triangle, $1$ at the top widening to $25$ at the base. The general formula for the sum of the first $n$ natural numbers,

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

is the same rule written out, and it is worth committing to memory — it reappears in Questions 9 and 6 of the End-of-Chapter Exercises.

Common mistakes

• Using $a + nd$ instead of $a + (n-1)d$. Reaching the $10$th term takes nine steps, not ten. This single slip accounts for more wrong AP answers than everything else combined.
• Question 1, checking only one difference. A sequence whose first two gaps match but whose third does not is not an AP at all. Check at least two gaps before using the formula.
• Question 2, answering “no” for $0$. It is tempting to think $0$ cannot be a term of a list that starts at $21$, but the AP is decreasing and passes straight through it at $n = 8$.
• Question 4, taking “last term” as $t_{49}$. A $50$-term AP has $t_{50}$ as its last term, so the equation is $a + 49d = 106$.
• Question 5, starting at $10$ or $9$. The first two-digit multiple of $3$ is $12$; $9$ is not a two-digit number and $10$ is not a multiple of $3$.
• Question 6, answering “$11$ years”. $n = 11$ is the year number, not the number of years that have passed. Ten increments get him there.
• Question 7, multiplying instead of adding. $25 \times 25$ would be the answer if every row had $25$ marbles. The rows differ, so they must be summed.

Practise next

• Exercise Set 8.3 — geometric progressions, where each term is multiplied rather than increased by a fixed amount.
• The End-of-Chapter Exercises, Questions 1 to 4 and 8 — five more AP questions of exactly this kind, two of them counting multiples in a range.