NCERT Class 10 Mathematics — Arithmetic Progressions, Exercise 5.3. All 20 questions solved.

Exercise 5.3 is the sum exercise. Exercise 5.2 asked for one term at a time; here the question is what a whole run of terms adds up to, or, turned around, how many terms or what first term produce a given total. For an AP with first term $a$ and common difference $d$, three formulae do all the work:

$$a_n = a + (n-1)d$$ $$S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]$$ $$S_n = \frac{n}{2}\,(a + l)$$

Here $l = a_n$ is the last of the $n$ terms. The third formula is the second with $2a + (n-1)d$ rewritten as $a + a_n$, and it is the quicker one whenever the last term is known. Question 11 adds a fact that runs the other way: once you know the sums, $a_n = S_n – S_{n-1}$.

Key insight. Every AP problem involves five quantities, $a$, $d$, $n$, $a_n$ and $S_n$. The two formulae $a_n = a + (n-1)d$ and $S_n = \frac{n}{2}[2a + (n-1)d]$ are two equations linking them, so any three of the five fix the other two. Every question below, from the ten parts of question 3 to the logs and potatoes at the end, comes down to one move: see which three are given, pick the formula that contains them, and solve. When $n$ comes out of a quadratic, test both roots against what $n$ means.

Question 1

Find the sum of the following APs:

(i) $2, 7, 12, \ldots$, to 10 terms.    (ii) $-37, -33, -29, \ldots$, to 12 terms.

(iii) $0.6, 1.7, 2.8, \ldots$, to 100 terms.    (iv) $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, \ldots$, to 11 terms.

Solution. In each part, read $a$ and $d$ off the first two terms. The last term is not given, so use $S_n = \frac{n}{2}[2a + (n-1)d]$.

(i) $a = 2$, $d = 5$, $n = 10$:

$$S_{10} = \frac{10}{2}\bigl[2(2) + 9(5)\bigr] = 5(4 + 45) = 245$$

(ii) $a = -37$, $d = -33 – (-37) = 4$, $n = 12$:

$$S_{12} = \frac{12}{2}\bigl[2(-37) + 11(4)\bigr] = 6(-74 + 44) = -180$$

The sum is negative because the first ten terms are all negative (the tenth is $-37 + 9 \times 4 = -1$), and the last two terms, $3$ and $7$, are far too small to cancel them.

(iii) $a = 0.6$, $d = 1.7 – 0.6 = 1.1$, $n = 100$:

$$S_{100} = \frac{100}{2}\bigl[1.2 + 99(1.1)\bigr] = 50(1.2 + 108.9) = 50 \times 110.1 = 5505$$

(iv) Fractions change nothing. Over the common denominator $60$ the terms are $\frac{4}{60}, \frac{5}{60}, \frac{6}{60}, \ldots$, so $a = \frac{1}{15}$ and $d = \frac{1}{60}$:

$$S_{11} = \frac{11}{2}\Bigl[\frac{2}{15} + 10 \cdot \frac{1}{60}\Bigr] = \frac{11}{2}\Bigl[\frac{8}{60} + \frac{10}{60}\Bigr] = \frac{11}{2} \cdot \frac{3}{10} = \frac{33}{20}$$

Question 2

Find the sums given below:

(i) $7 + 10\tfrac{1}{2} + 14 + \ldots + 84$    (ii) $34 + 32 + 30 + \ldots + 10$

(iii) $-5 + (-8) + (-11) + \ldots + (-230)$

Solution. This time the last term is given but the number of terms is not. So find $n$ first from $a_n = a + (n-1)d$, and then use $S_n = \frac{n}{2}(a + l)$, which needs nothing else.

(i) $a = 7$, $d = 10\frac{1}{2} – 7 = \frac{7}{2}$, $l = 84$:

$$84 = 7 + (n-1)\cdot\frac{7}{2} \;\Rightarrow\; n – 1 = 22 \;\Rightarrow\; n = 23$$

$$S_{23} = \frac{23}{2}(7 + 84) = \frac{23 \times 91}{2} = \frac{2093}{2} = 1046\tfrac{1}{2}$$

(ii) $a = 34$, $d = -2$, $l = 10$:

$$10 = 34 + (n-1)(-2) \;\Rightarrow\; n – 1 = 12 \;\Rightarrow\; n = 13$$

$$S_{13} = \frac{13}{2}(34 + 10) = 13 \times 22 = 286$$

(iii) $a = -5$, $d = -3$, $l = -230$:

$$-230 = -5 + (n-1)(-3) \;\Rightarrow\; n – 1 = 75 \;\Rightarrow\; n = 76$$

$$S_{76} = \frac{76}{2}\bigl(-5 + (-230)\bigr) = 38 \times (-235) = -8930$$

Question 3

In an AP:

(i) given $a = 5$, $d = 3$, $a_n = 50$, find $n$ and $S_n$.

(ii) given $a = 7$, $a_{13} = 35$, find $d$ and $S_{13}$.

(iii) given $a_{12} = 37$, $d = 3$, find $a$ and $S_{12}$.

(iv) given $a_3 = 15$, $S_{10} = 125$, find $d$ and $a_{10}$.

(v) given $d = 5$, $S_9 = 75$, find $a$ and $a_9$.

(vi) given $a = 2$, $d = 8$, $S_n = 90$, find $n$ and $a_n$.

(vii) given $a = 8$, $a_n = 62$, $S_n = 210$, find $n$ and $d$.

(viii) given $a_n = 4$, $d = 2$, $S_n = -14$, find $n$ and $a$.

(ix) given $a = 3$, $n = 8$, $S = 192$, find $d$.

(x) given $l = 28$, $S = 144$, and there are total 9 terms. Find $a$.

Solution. Each part hands over three of the five quantities. The skill is choosing the formula that uses them directly.

(i) The $n$th term formula contains $a$, $d$ and $a_n$, so it gives $n$:

$$50 = 5 + (n-1)3 \;\Rightarrow\; n – 1 = 15 \;\Rightarrow\; n = 16$$

Now the first and last terms are both known, so

$$S_{16} = \frac{16}{2}(5 + 50) = 8 \times 55 = 440$$

(ii) $a_{13} = a + 12d$, so $7 + 12d = 35$ and $d = \dfrac{28}{12} = \dfrac{7}{3}$. With the first and thirteenth terms known,

$$S_{13} = \frac{13}{2}(7 + 35) = 13 \times 21 = 273$$

(iii) $a_{12} = a + 11d$, so $a + 33 = 37$ and $a = 4$. Then

$$S_{12} = \frac{12}{2}(4 + 37) = 6 \times 41 = 246$$

(iv) Two unknowns, $a$ and $d$, need two equations, and the question supplies exactly two facts:

$$a_3 = a + 2d = 15, \qquad S_{10} = \frac{10}{2}(2a + 9d) = 125 \;\Rightarrow\; 2a + 9d = 25$$

Doubling the first gives $2a + 4d = 30$. Subtracting it from the second leaves $5d = -5$, so $d = -1$ and $a = 15 – 2d = 17$. Therefore

$$a_{10} = a + 9d = 17 – 9 = 8$$

(v) Only $a$ is unknown in the sum formula:

$$S_9 = \frac{9}{2}\bigl[2a + 8(5)\bigr] = 75 \;\Rightarrow\; 2a + 40 = \frac{50}{3} \;\Rightarrow\; a = -\frac{35}{3}$$

$$a_9 = a + 8d = -\frac{35}{3} + 40 = \frac{85}{3}$$

A fractional first term is nothing to worry about; nothing in the question says the terms must be whole numbers.

(vi) Now $n$ is the unknown in the sum formula, and it appears twice, so expect a quadratic:

$$\frac{n}{2}\bigl[2(2) + (n-1)8\bigr] = 90 \;\Rightarrow\; n(4n – 2) = 90 \;\Rightarrow\; 2n^2 – n – 45 = 0$$

$$(n – 5)(2n + 9) = 0 \;\Rightarrow\; n = 5 \text{ or } n = -\tfrac{9}{2}$$

A number of terms must be a positive integer, so $n = 5$, and $a_5 = 2 + 4 \times 8 = 34$.

(vii) First and last terms are given, so the second sum formula gives $n$ at once:

$$S_n = \frac{n}{2}(8 + 62) = 35n = 210 \;\Rightarrow\; n = 6$$

$$a_6 = 8 + 5d = 62 \;\Rightarrow\; d = \frac{54}{5}$$

(viii) Here $a$ and $n$ are both unknown. Express $a$ through $n$ using the last term, $4 = a + (n-1)2$, so $a = 6 – 2n$. Then

$$S_n = \frac{n}{2}(a + a_n) = \frac{n}{2}(6 – 2n + 4) = n(5 – n) = -14$$

$$n^2 – 5n – 14 = 0 \;\Rightarrow\; (n – 7)(n + 2) = 0 \;\Rightarrow\; n = 7$$

rejecting $n = -2$. So $a = 6 – 14 = -8$. The AP is $-8, -6, -4, -2, 0, 2, 4$, and those seven terms do add up to $-14$.

(ix) $S_8 = \dfrac{8}{2}\bigl[2(3) + 7d\bigr] = 192$, so $6 + 7d = 48$ and $d = 6$.

(x) The last term is $l = 28$ and $n = 9$, so

$$\frac{9}{2}(a + 28) = 144 \;\Rightarrow\; a + 28 = 32 \;\Rightarrow\; a = 4$$

Question 4

How many terms of the AP: $9, 17, 25, \ldots$ must be taken to give a sum of $636$?

Solution. Here $a = 9$ and $d = 8$, and the unknown is $n$:

$$\frac{n}{2}\bigl[2(9) + (n-1)8\bigr] = 636 \;\Rightarrow\; n(4n + 5) = 636 \;\Rightarrow\; 4n^2 + 5n – 636 = 0$$

The discriminant is $25 + 16 \times 636 = 10201 = 101^2$, so

$$n = \frac{-5 \pm 101}{8} = 12 \quad\text{or}\quad -\frac{53}{4}$$

The negative fraction cannot count terms, so $n = 12$. As a check, $a_{12} = 9 + 88 = 97$ and $S_{12} = 6(9 + 97) = 636$.

Question 5

The first term of an AP is $5$, the last term is $45$ and the sum is $400$. Find the number of terms and the common difference.

Solution. First term, last term and sum: exactly the three quantities in $S_n = \frac{n}{2}(a + l)$.

$$400 = \frac{n}{2}(5 + 45) = 25n \;\Rightarrow\; n = 16$$

The sixteenth term is the last one, so $45 = 5 + 15d$, giving $d = \dfrac{40}{15} = \dfrac{8}{3}$.

Question 6

The first and the last terms of an AP are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there and what is their sum?

Solution. Find $n$ from the last term:

$$350 = 17 + (n-1)9 \;\Rightarrow\; n – 1 = 37 \;\Rightarrow\; n = 38$$

$$S_{38} = \frac{38}{2}(17 + 350) = 19 \times 367 = 6973$$

Question 7

Find the sum of first $22$ terms of an AP in which $d = 7$ and $22$nd term is $149$.

Solution. The sum formula $\frac{n}{2}(a + l)$ needs the first term, and the 22nd term supplies it: $149 = a + 21 \times 7$, so $a = 2$. Then

$$S_{22} = \frac{22}{2}(2 + 149) = 11 \times 151 = 1661$$

Question 8

Find the sum of first $51$ terms of an AP whose second and third terms are $14$ and $18$ respectively.

Solution. Consecutive terms differ by $d$, so $d = 18 – 14 = 4$, and stepping back once from the second term gives $a = 14 – 4 = 10$.

$$S_{51} = \frac{51}{2}\bigl[2(10) + 50(4)\bigr] = \frac{51}{2} \times 220 = 5610$$

Question 9

If the sum of first $7$ terms of an AP is $49$ and that of $17$ terms is $289$, find the sum of first $n$ terms.

Solution. The answer has to be a formula in $n$, which means finding $a$ and $d$ first. Each given sum is one equation:

$$S_7 = \frac{7}{2}(2a + 6d) = 49 \;\Rightarrow\; a + 3d = 7$$

$$S_{17} = \frac{17}{2}(2a + 16d) = 289 \;\Rightarrow\; a + 8d = 17$$

Subtracting, $5d = 10$, so $d = 2$ and $a = 1$. The AP is $1, 3, 5, \ldots$, the odd numbers, and

$$S_n = \frac{n}{2}\bigl[2 + (n-1)2\bigr] = \frac{n}{2} \cdot 2n = n^2$$

Checking against the data: $7^2 = 49$ and $17^2 = 289$.

Question 10

Show that $a_1, a_2, \ldots, a_n, \ldots$ form an AP where $a_n$ is defined as below:

(i) $a_n = 3 + 4n$    (ii) $a_n = 9 – 5n$

Also find the sum of the first $15$ terms in each case.

Solution. A sequence is an AP when the difference between consecutive terms is the same for every $n$. So compute $a_{n+1} – a_n$ in general, not just for the first few terms.

(i) $a_{n+1} – a_n = [3 + 4(n+1)] – [3 + 4n] = 4$, which does not depend on $n$, so the sequence is an AP with $d = 4$. Its first term is $a_1 = 7$ and its fifteenth is $a_{15} = 63$:

$$S_{15} = \frac{15}{2}(7 + 63) = 525$$

(ii) $a_{n+1} – a_n = [9 – 5(n+1)] – [9 – 5n] = -5$, again constant, so this is an AP with $d = -5$. Here $a_1 = 4$ and $a_{15} = 9 – 75 = -66$:

$$S_{15} = \frac{15}{2}\bigl(4 + (-66)\bigr) = \frac{15}{2}(-62) = -465$$

The pattern is general: any $a_n$ that is linear in $n$ gives an AP, and the coefficient of $n$ is the common difference.

Question 11

If the sum of the first $n$ terms of an AP is $4n – n^2$, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the $3$rd, the $10$th and the $n$th terms.

Solution. The key fact is that $S_n$ and $S_{n-1}$ differ by exactly one term, the $n$th:

$$a_n = S_n – S_{n-1}$$

Using $S_n = 4n – n^2$:

• $S_1 = 4 – 1 = 3$. The sum of one term is that term, so $a_1 = 3$.
• $S_2 = 8 – 4 = 4$, so $a_2 = S_2 – S_1 = 4 – 3 = 1$.
• $S_3 = 12 – 9 = 3$, so $a_3 = S_3 – S_2 = 3 – 4 = -1$.
• $S_{10} = 40 – 100 = -60$ and $S_9 = 36 – 81 = -45$, so $a_{10} = -60 – (-45) = -15$.

In general,

$$a_n = \bigl[4n – n^2\bigr] – \bigl[4(n-1) – (n-1)^2\bigr] = 4 – (2n – 1) = 5 – 2n$$

This agrees with every value above: $5 – 2 = 3$, $5 – 4 = 1$, $5 – 6 = -1$, $5 – 20 = -15$.

Question 12

Find the sum of the first $40$ positive integers divisible by $6$.

Solution. The positive integers divisible by $6$ are $6, 12, 18, \ldots$, an AP with $a = d = 6$. The fortieth is $6 \times 40 = 240$.

$$S_{40} = \frac{40}{2}(6 + 240) = 20 \times 246 = 4920$$

Question 13

Find the sum of the first $15$ multiples of $8$.

Solution. The multiples $8, 16, \ldots, 120$ form an AP with $a = d = 8$ and $l = 8 \times 15 = 120$:

$$S_{15} = \frac{15}{2}(8 + 120) = 15 \times 64 = 960$$

Question 14

Find the sum of the odd numbers between $0$ and $50$.

Solution. The odd numbers are $1, 3, 5, \ldots, 49$, with $a = 1$, $d = 2$ and $l = 49$. The count comes from the last term, $49 = 1 + (n-1)2$, so $n = 25$:

$$S_{25} = \frac{25}{2}(1 + 49) = 625$$

This is $25^2$, as question 9 predicts: the sum of the first $n$ odd numbers is $n^2$.

Question 15

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ $200$ for the first day, ₹ $250$ for the second day, ₹ $300$ for the third day, etc., the penalty for each succeeding day being ₹ $50$ more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by $30$ days?

Solution. The daily penalties form an AP with $a = 200$ and $d = 50$. The total for $30$ days is the sum of its first $30$ terms:

$$S_{30} = \frac{30}{2}\bigl[2(200) + 29(50)\bigr] = 15(400 + 1450) = 15 \times 1850 = 27750$$

Question 16

A sum of ₹ $700$ is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ $20$ less than its preceding prize, find the value of each of the prizes.

Solution. Taking the largest prize as $a$, the prizes form an AP with $d = -20$ and $n = 7$, and together they total $700$:

$$\frac{7}{2}\bigl[2a + 6(-20)\bigr] = 700 \;\Rightarrow\; 2a – 120 = 200 \;\Rightarrow\; a = 160$$

The prizes are then found by subtracting $20$ each time. They add up to $700$, as they should.

Question 17

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant $1$ tree, a section of Class II will plant $2$ trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution. With three sections per class, Class I plants $3 \times 1 = 3$ trees, Class II plants $3 \times 2 = 6$, and so on up to Class XII with $3 \times 12 = 36$. That is an AP with $a = 3$, $d = 3$, $n = 12$:

$$S_{12} = \frac{12}{2}(3 + 36) = 6 \times 39 = 234$$

Question 18

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii $0.5$ cm, $1.0$ cm, $1.5$ cm, $2.0$ cm, $\ldots$ as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = \dfrac{22}{7}$)

[Hint: Length of successive semicircles is $l_1, l_2, l_3, l_4, \ldots$ with centres at A, B, A, B, $\ldots$, respectively.]

In Fig. 5.4, A and B lie on a horizontal line, $0.5$ cm apart. The first semicircle, centred at A, runs over the top from B; the second, centred at B, runs underneath; and so on, each one starting where the last one ended:

Solution. A semicircle of radius $r$ has length $\pi r$, half the circumference $2\pi r$. The radii go up by $0.5$ cm each time, so the thirteenth semicircle has radius $13 \times 0.5 = 6.5$ cm, and

$$l_1 + l_2 + \cdots + l_{13} = \pi(0.5 + 1.0 + 1.5 + \cdots + 6.5)$$

The bracket is an AP with $a = 0.5$, $l = 6.5$ and $n = 13$:

$$0.5 + 1.0 + \cdots + 6.5 = \frac{13}{2}(0.5 + 6.5) = \frac{13 \times 7}{2} = \frac{91}{2}$$

$$\text{Total length} = \frac{22}{7} \times \frac{91}{2} = 11 \times 13 = 143 \text{ cm}$$

Question 19

$200$ logs are stacked in the following manner: $20$ logs in the bottom row, $19$ in the next row, $18$ in the row next to it and so on (see Fig. 5.5). In how many rows are the $200$ logs placed and how many logs are in the top row?

Fig. 5.5 shows a pile of logs seen end-on, each row one log shorter than the row beneath it.

Solution. Counting from the bottom, the rows form an AP with $a = 20$ and $d = -1$. The number of rows $n$ is the number of terms whose sum is $200$:

$$\frac{n}{2}\bigl[2(20) + (n-1)(-1)\bigr] = 200 \;\Rightarrow\; n(41 – n) = 400 \;\Rightarrow\; n^2 – 41n + 400 = 0$$

$$(n – 16)(n – 25) = 0 \;\Rightarrow\; n = 16 \text{ or } n = 25$$

Unlike questions 3(vi) and 4, both roots are positive whole numbers, so the equation alone cannot decide. The context has to. With $n = 25$ the top row would hold $a_{25} = 20 – 24 = -4$ logs, which is impossible. (Algebraically, rows 17 to 25 would hold $4, 3, 2, 1, 0, -1, -2, -3, -4$ logs, which add up to zero, and that is why the sum comes back to $200$.) With $n = 16$, the top row holds $a_{16} = 20 – 15 = 5$ logs.

Question 20

In a potato race, a bucket is placed at the starting point, which is $5$ m from the first potato, and the other potatoes are placed $3$ m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is $2 \times 5 + 2 \times (5 + 3)$]

Solution. Every potato is fetched from the bucket, so what matters is each potato’s distance from the bucket: $5, 8, 11, \ldots$ metres, an AP with $a = 5$, $d = 3$ and last term $5 + 9 \times 3 = 32$. Each potato costs a run out and a run back, so twice its distance:

$$\text{Total} = 2(5 + 8 + 11 + \cdots + 32) = 2 \times \frac{10}{2}(5 + 32) = 2 \times 185 = 370 \text{ m}$$

This is the hint’s pattern continued: the round trips $10, 16, 22, \ldots$ are themselves an AP with $a = 10$, $d = 6$, and $S_{10} = 5(20 + 54) = 370$.

Common mistakes

• Miscounting the terms in question 2. The number of terms is $\frac{l – a}{d} + 1$, not $\frac{l – a}{d}$. In question 2(iii), $\frac{-230 – (-5)}{-3} = 75$ counts the gaps between terms, and forgetting the $+1$ gives $75$ terms instead of $76$ and a sum of $-8812.5$ instead of $-8930$.
• Getting $d$ wrong in question 1(iv). $\frac{1}{12} – \frac{1}{15}$ is $\frac{5 – 4}{60} = \frac{1}{60}$. It is not $\frac{1}{3}$, which comes from subtracting the denominators. Rewriting the terms as $\frac{4}{60}, \frac{5}{60}, \frac{6}{60}$ makes the common difference obvious.
• Keeping the wrong root. In questions 3(vi), 3(viii) and 4 the rejected root is negative or a fraction, so it is easy to discard. Question 19 is the trap: both $16$ and $25$ are positive integers, and only checking the top row ($-4$ logs) rules out $25$.
• Treating $S_n$ as $a_n$ in question 11. $4n – n^2$ is the sum of the first $n$ terms, not the $n$th term. Substituting $n = 10$ gives $S_{10} = -60$; the tenth term is $S_{10} – S_9 = -15$.
• Using the full circumference in question 18. Each piece of the spiral is a semicircle, of length $\pi r$. Using $2\pi r$ doubles the answer to $286$ cm.
• Forgetting the return trips in question 20. Adding only the distances to the potatoes gives $185$ m. The competitor has to bring every potato back, so each distance is run twice. Measuring each potato from the previous potato, rather than from the bucket, is the other common slip.
• Miscounting “between 0 and 50” in question 14. The odd numbers run from $1$ to $49$, which is $25$ numbers, not $24$ or $26$. Find $n$ from $49 = 1 + (n-1)2$ rather than by guessing.

Practise next

• Exercise 5.4 (Optional) — harder problems where the AP has to be found inside the situation first: rungs of a ladder, house numbers and the steps of a terrace.
• Exercise 5.2 — the $n$th term formula $a_n = a + (n-1)d$, which every question here uses to find $n$, $a$ or $d$ before the sum.
• Chapter 4, Exercise 4.2 — solving quadratics by factorisation, the step that decides $n$ in questions 3(vi), 3(viii), 4 and 19.