NCERT Class 10 Mathematics — Arithmetic Progressions, Exercise 5.1. All 4 questions solved.

Exercise 5.1 is about recognising an arithmetic progression and writing one down. It uses a single definition.

An arithmetic progression (AP) is a list of numbers in which each term after the first is obtained by adding a fixed number, the common difference $d$, to the term before it. With first term $a$, an AP is

$$a,\quad a + d,\quad a + 2d,\quad a + 3d,\ \ldots$$

and the test for a given list $a_1, a_2, a_3, \ldots$ is that every difference $a_{k+1} – a_k$ is the same number. That number is $d$, and it may be positive, negative or zero.

Key insight. An AP is recognised by its differences, never by the look of its terms. Work out every consecutive difference you are given. If even one disagrees, the list is not an AP; if they all agree, their common value is $d$. Question 4 is built to test exactly this. Lists (xiii) and (xiv) look neatly patterned and fail the test. List (xv) mixes squares with a plain $73$, looks irregular, and passes.

Question 1

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is ₹ $15$ for the first km and ₹ $8$ for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes $\dfrac{1}{4}$ of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs ₹ $150$ for the first metre and rises by ₹ $50$ for each subsequent metre.

(iv) The amount of money in the account every year, when ₹ $10000$ is deposited at compound interest at $8\%$ per annum.

Solution. Write out the first few numbers of each list, then check the differences.

(i) After $1$ km the fare is ₹ $15$; each further km adds ₹ $8$. The list is

$$15,\ 23,\ 31,\ 39,\ \ldots$$

Every term is the previous one plus $8$, so it is an AP, with $a = 15$ and $d = 8$.

(ii) Let the cylinder start with volume $V$ of air. Each stroke removes a quarter of what is left, so leaves three-quarters of it:

$$V,\quad \frac{3V}{4},\quad \left(\frac{3}{4}\right)^2 V = \frac{9V}{16},\quad \left(\frac{3}{4}\right)^3 V = \frac{27V}{64},\ \ldots$$

The differences are $-\dfrac{V}{4}$, then $-\dfrac{3V}{16}$, then $-\dfrac{9V}{64}$. They shrink, because a quarter of a smaller amount is a smaller amount. Not an AP. (Each term is a fixed multiple, $\frac{3}{4}$, of the one before, which makes it a geometric progression instead.)

(iii) The cost is ₹ $150$ for the first metre and rises by ₹ $50$ with each further metre:

$$150,\ 200,\ 250,\ 300,\ \ldots$$

A constant ₹ $50$ is added each time, so this is an AP with $a = 150$ and $d = 50$.

The wording can also be read as a running total of the whole cost so far: $150$, then $150 + 200 = 350$, then $350 + 250 = 600$. That list is not an AP. The textbook’s answer takes the first reading, which is the one followed here.

(iv) Compound interest adds $8\%$ of the current amount each year, so each year’s amount is $1.08$ times the last:

$$10000\left(1 + \frac{8}{100}\right),\quad 10000\left(1 + \frac{8}{100}\right)^2,\quad 10000\left(1 + \frac{8}{100}\right)^3,\ \ldots$$

that is, $10800,\ 11664,\ 12597.12,\ \ldots$ The differences are $864$ and $933.12$. The interest grows because it is earned on a growing balance. Not an AP. (Under simple interest the same ₹ $800$ would be added every year, and the list would be an AP.)

Question 2

Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:

(i) $a = 10$, $d = 10$    (ii) $a = -2$, $d = 0$    (iii) $a = 4$, $d = -3$

(iv) $a = -1$, $d = \dfrac{1}{2}$    (v) $a = -1.25$, $d = -0.25$

Solution. The terms are $a,\ a + d,\ a + 2d,\ a + 3d$. In other words, start at $a$ and keep adding $d$.

(i) $10,\ 20,\ 30,\ 40$

(ii) Adding $0$ changes nothing, so every term is $-2$: $-2,\ -2,\ -2,\ -2$. A constant list is still an AP, with $d = 0$.

(iii) Adding $-3$ means subtracting $3$: $4,\ 1,\ -2,\ -5$.

(iv) $-1,\ -\dfrac{1}{2},\ 0,\ \dfrac{1}{2}$

(v) Each step goes a quarter further below zero: $-1.25,\ -1.50,\ -1.75,\ -2.00$.

Question 3

For the following APs, write the first term and the common difference:

(i) $3, 1, -1, -3, \ldots$    (ii) $-5, -1, 3, 7, \ldots$

(iii) $\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{13}{3}, \ldots$    (iv) $0.6, 1.7, 2.8, 3.9, \ldots$

Solution. The first term is simply $a_1$. The common difference is any term minus the one before it, $d = a_2 – a_1$. Checking a second pair confirms it.

(i) $a = 3$ and $d = 1 – 3 = -2$ (and $-1 – 1 = -2$ again).

(ii) $a = -5$ and $d = -1 – (-5) = 4$.

(iii) $a = \dfrac{1}{3}$ and $d = \dfrac{5}{3} – \dfrac{1}{3} = \dfrac{4}{3}$.

(iv) $a = 0.6$ and $d = 1.7 – 0.6 = 1.1$.

Question 4

Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms.

(i) $2, 4, 8, 16, \ldots$    (ii) $2, \dfrac{5}{2}, 3, \dfrac{7}{2}, \ldots$    (iii) $-1.2, -3.2, -5.2, -7.2, \ldots$

(iv) $-10, -6, -2, 2, \ldots$    (v) $3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \ldots$

(vi) $0.2, 0.22, 0.222, 0.2222, \ldots$    (vii) $0, -4, -8, -12, \ldots$

(viii) $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, \ldots$    (ix) $1, 3, 9, 27, \ldots$

(x) $a, 2a, 3a, 4a, \ldots$    (xi) $a, a^2, a^3, a^4, \ldots$

(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$    (xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$

(xiv) $1^2, 3^2, 5^2, 7^2, \ldots$    (xv) $1^2, 5^2, 7^2, 73, \ldots$

Solution. For each list, compute all three differences between the four given terms. Where they agree, keep adding $d$ to the last term to get three more.

(i) Differences $2, 4, 8$. Each term doubles instead. Not an AP.

(ii) Differences $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$. AP, $d = \frac{1}{2}$. Next three: $4,\ \frac{9}{2},\ 5$.

(iii) Differences $-2, -2, -2$. AP, $d = -2$. Next three: $-9.2,\ -11.2,\ -13.2$.

(iv) Differences $4, 4, 4$. AP, $d = 4$. Next three: $6,\ 10,\ 14$.

(v) Each term adds one more $\sqrt{2}$, so the differences are $\sqrt{2}, \sqrt{2}, \sqrt{2}$. AP, $d = \sqrt{2}$. Next three: $3 + 4\sqrt{2},\ 3 + 5\sqrt{2},\ 3 + 6\sqrt{2}$.

(vi) Differences $0.02, 0.002, 0.0002$. Each extra digit adds ten times less than the one before. Not an AP.

(vii) Differences $-4, -4, -4$. AP, $d = -4$. Next three: $-16,\ -20,\ -24$.

(viii) Differences $0, 0, 0$. AP, $d = 0$. Next three: $-\frac{1}{2},\ -\frac{1}{2},\ -\frac{1}{2}$.

(ix) Differences $2, 6, 18$. Each term triples. Not an AP.

(x) Differences $a, a, a$. AP, $d = a$. Next three: $5a,\ 6a,\ 7a$.

(xi) Differences $a^2 – a = a(a – 1)$, then $a^3 – a^2 = a^2(a – 1)$, then $a^3(a – 1)$. Each is $a$ times the one before, so they agree only in the special cases $a = 0$ and $a = 1$ (when the list is constant). For a general $a$ it is not an AP.

(xii) Simplify the surds first: $\sqrt{8} = 2\sqrt{2}$, $\sqrt{18} = 3\sqrt{2}$, $\sqrt{32} = 4\sqrt{2}$. The list is $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}$, so it is an AP with $d = \sqrt{2}$. Next three: $5\sqrt{2} = \sqrt{50},\ 6\sqrt{2} = \sqrt{72},\ 7\sqrt{2} = \sqrt{98}$.

(xiii) The numbers under the roots, $3, 6, 9, 12$, are in AP, but the roots themselves are not. The differences are $\sqrt{6} – \sqrt{3} \approx 0.717$, then $3 – \sqrt{6} \approx 0.551$, then $2\sqrt{3} – 3 \approx 0.464$. Not an AP.

(xiv) The list is $1, 9, 25, 49$, with differences $8, 16, 24$. Not an AP.

(xv) The list is $1, 25, 49, 73$, with differences $24, 24, 24$. It is an AP, with $d = 24$. Next three: $97,\ 121,\ 145$.

Common mistakes

• Question 1(ii), confusing “a fixed fraction” with “a fixed amount”. The pump removes the same fraction each time, but a quarter of less air is less air. An AP needs the same amount added or removed each time.
• Question 1(iv), treating compound interest like simple interest. Simple interest adds ₹ $800$ every year and gives an AP. Compound interest charges interest on interest, so the yearly increase itself grows.
• Question 3(i), subtracting the wrong way round. $d$ is a term minus the one before it: $1 – 3 = -2$. Writing $3 – 1 = 2$ describes an increasing list, which this is not.
• Question 4(viii) and question 2(ii), deciding a constant list is not an AP. Nothing in the definition stops $d$ being $0$. A list that never changes adds $0$ each time and is an AP.
• Question 4(xii) and (xiii), judging surds by the numbers inside. In (xii) the surds simplify to multiples of $\sqrt{2}$, and that is why the list is an AP. In (xiii) the radicands are in AP but the roots are not, because taking square roots does not preserve equal gaps.
• Question 4(xv), dismissing it because it mixes squares with a plain number. Evaluate first: $1, 25, 49, 73$ has a constant gap of $24$. The look of the terms is irrelevant; only the differences matter.

Practise next

• Exercise 5.2 — the $n$th term formula $a_n = a + (n – 1)d$, which finds any term directly instead of adding $d$ one step at a time as in question 4.
• Class 11 Sequences and Series, Exercise 8.2 — geometric progressions, the natural home of the lists that failed here by multiplying: question 1(ii) and (iv), and question 4(i) and (ix).