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NCERT Solutions · Class 10 Maths · Chapter 5

Chapter 5: Arithmetic Progressions (Algebra)

Step-by-step NCERT solutions for Class 10 Maths Chapter 5, Arithmetic Progressions (2026-27 reprint): Exercise 5.1 identifying APs, Exercise 5.2 the nth term, Exercise 5.3 the sum of n terms with word problems, and the optional Exercise 5.4. All 49 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-5-arithmetic-progressions

An AP is $a, a + d, a + 2 d, \dots$ The $n$ th term is $a_{n} = a + (n - 1) d$ . The sum of the first $n$ terms is $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l)$ , where $l$ is the last term.

Exercise 5.1

In which situations do the numbers form an AP, and why? (i) Taxi fare after each km: ₹15 for the first km and ₹8 for each additional km. (ii) The air in a cylinder when a pump removes 1/4 of the remaining air each time. (iii) The cost of digging a well after each metre: ₹150 for the first metre, rising by ₹50 for each further metre. (iv) Money in an account each year when ₹10000 earns 8% compound interest.

Solution

Fares: 15, 23, 31, 39, … Each term is 8 more than the previous one, so it is an AP ( $d = 8$ ).
Air left: $V, \frac{3}{4} V, \frac{9}{16} V, \dots$ The differences $- \frac{V}{4}, - \frac{3 V}{16}, \dots$ are not equal, so it is not an AP.
Costs: 150, 200, 250, … The difference is always 50, so it is an AP.
Amounts: $10000 (1.08), 10000 (1.08)^{2}, 10000 (1.08)^{3}, \dots$ Each is multiplied (not increased) by a fixed number, so the differences grow and it is not an AP.

(i) AP (d = 8) (ii) Not an AP (iii) AP (d = 50) (iv) Not an AP

Write the first four terms of the AP when (i) a = 10, d = 10 (ii) a = –2, d = 0 (iii) a = 4, d = –3 (iv) a = –1, d = 1/2 (v) a = –1.25, d = –0.25

Solution

10, 20, 30, 40
–2, –2, –2, –2
4, 1, –2, –5
$- 1, - \frac{1}{2}, 0, \frac{1}{2}$
–1.25, –1.50, –1.75, –2.00

(i) 10, 20, 30, 40 (ii) –2, –2, –2, –2 (iii) 4, 1, –2, –5 (iv) –1, –1/2, 0, 1/2 (v) –1.25, –1.50, –1.75, –2.00

Write the first term and common difference: (i) 3, 1, –1, –3, … (ii) –5, –1, 3, 7, … (iii) 1/3, 5/3, 9/3, 13/3, … (iv) 0.6, 1.7, 2.8, 3.9, …

Solution

$a = 3$ , $d = 1 - 3 = - 2$
$a = - 5$ , $d = 4$
$a = \frac{1}{3}$ , $d = \frac{4}{3}$
$a = 0.6$ , $d = 1.1$

(i) a = 3, d = –2 (ii) a = –5, d = 4 (iii) a = 1/3, d = 4/3 (iv) a = 0.6, d = 1.1

Which are APs? For an AP, find d and three more terms. (i) 2, 4, 8, 16, … (ii) 2, 5/2, 3, 7/2, … (iii) –1.2, –3.2, –5.2, –7.2, … (iv) –10, –6, –2, 2, … (v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, … (vi) 0.2, 0.22, 0.222, 0.2222, … (vii) 0, –4, –8, –12, … (viii) –1/2, –1/2, –1/2, … (ix) 1, 3, 9, 27, … (x) a, 2a, 3a, 4a, … (xi) a, a², a³, a⁴, … (xii) √2, √8, √18, √32, … (xiii) √3, √6, √9, √12, … (xiv) 1², 3², 5², 7², … (xv) 1², 5², 7², 73, …

Solution

Check whether every difference is the same.

	AP?	d	Next three terms
(i)	No (differences 2, 4, 8)
(ii)	Yes	1/2	4, 9/2, 5
(iii)	Yes	–2	–9.2, –11.2, –13.2
(iv)	Yes	4	6, 10, 14
(v)	Yes	√2	3 + 4√2, 3 + 5√2, 3 + 6√2
(vi)	No (differences 0.02, 0.002, …)
(vii)	Yes	–4	–16, –20, –24
(viii)	Yes	0	–1/2, –1/2, –1/2
(ix)	No (differences 2, 6, 18)
(x)	Yes	a	5a, 6a, 7a
(xi)	No (differences a² – a, a³ – a², … are unequal in general)
(xii)	Yes: the terms are √2, 2√2, 3√2, 4√2	√2	5√2, 6√2, 7√2 (= √50, √72, √98)
(xiii)	No (√6 – √3 ≠ √9 – √6)
(xiv)	No: 1, 9, 25, 49 (differences 8, 16, 24)
(xv)	Yes: 1, 25, 49, 73	24	97, 121, 145

APs: (ii), (iii), (iv), (v), (vii), (viii), (x), (xii), (xv). Not APs: (i), (vi), (ix), (xi), (xiii), (xiv).

Exercise 5.2

Fill in the blanks (a = first term, d = common difference, aₙ = nth term): (i) a = 7, d = 3, n = 8 (ii) a = –18, n = 10, aₙ = 0 (iii) d = –3, n = 18, aₙ = –5 (iv) a = –18.9, d = 2.5, aₙ = 3.6 (v) a = 3.5, d = 0, n = 105

Solution

Use $a_{n} = a + (n - 1) d$ :

$a_{8} = 7 + 7 \times 3 = 28$
$0 = - 18 + 9 d \Rightarrow d = 2$
$- 5 = a + 17 (- 3) \Rightarrow a = 46$
$3.6 = - 18.9 + (n - 1) (2.5) \Rightarrow n - 1 = 9 \Rightarrow n = 10$
$a_{105} = 3.5 + 0 = 3.5$

(i) aₙ = 28 (ii) d = 2 (iii) a = 46 (iv) n = 10 (v) aₙ = 3.5

Choose the correct option and justify: (i) The 30th term of 10, 7, 4, … is (A) 97 (B) 77 (C) –77 (D) –87. (ii) The 11th term of –3, –1/2, 2, … is (A) 28 (B) 22 (C) –38 (D) –48½.

Solution

$a = 10$ , $d = - 3$ : $a_{30} = 10 + 29 (- 3) = - 77$ , option (C).
$a = - 3$ , $d = \frac{5}{2}$ : $a_{11} = - 3 + 10 \times \frac{5}{2} = 22$ , option (B).

(i) (C) –77 (ii) (B) 22

Find the missing terms: (i) 2, __, 26 (ii) __, 13, __, 3 (iii) 5, __, __, 9½ (iv) –4, __, __, __, __, 6 (v) __, 38, __, __, __, –22

Solution

$26 = 2 + 2 d \Rightarrow d = 12$ : the missing term is 14.
$a_{2} = 13$ , $a_{4} = 3$ : $2 d = - 10 \Rightarrow d = - 5$ : 18, 13, 8, 3.
$9 \frac{1}{2} = 5 + 3 d \Rightarrow d = \frac{3}{2}$ : 5, 6½, 8, 9½.
$6 = - 4 + 5 d \Rightarrow d = 2$ : –4, –2, 0, 2, 4, 6.
$a_{2} = 38$ , $a_{6} = - 22$ : $4 d = - 60 \Rightarrow d = - 15$ : 53, 38, 23, 8, –7, –22.

(i) 14 (ii) 18, 8 (iii) 6½, 8 (iv) –2, 0, 2, 4 (v) 53, 23, 8, –7

Which term of the AP 3, 8, 13, 18, … is 78?

Solution

$78 = 3 + (n - 1) 5 \Rightarrow n - 1 = 15 \Rightarrow n = 16$

The 16th term.

Find the number of terms: (i) 7, 13, 19, …, 205 (ii) 18, 15½, 13, …, –47

Solution

$205 = 7 + (n - 1) 6 \Rightarrow n - 1 = 33 \Rightarrow n = 34$
$d = - \frac{5}{2}$ : $- 47 = 18 + (n - 1) (- \frac{5}{2}) \Rightarrow n - 1 = 26 \Rightarrow n = 27$

(i) 34 terms (ii) 27 terms

Check whether –150 is a term of the AP 11, 8, 5, 2, …

Solution

$- 150 = 11 + (n - 1) (- 3) \Rightarrow n - 1 = \frac{161}{3}$ , which is not a whole number.

No, –150 is not a term.

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

Solution

$a + 10 d = 38$ and $a + 15 d = 73 \Rightarrow 5 d = 35 \Rightarrow d = 7$ , $a = - 32$ .

$a_{31} = - 32 + 30 \times 7 = 178$

178

An AP has 50 terms; the 3rd term is 12 and the last term is 106. Find the 29th term.

Solution

$a + 2 d = 12$ and $a + 49 d = 106 \Rightarrow 47 d = 94 \Rightarrow d = 2$ , $a = 8$ .

$a_{29} = 8 + 28 \times 2 = 64$

The 3rd and 9th terms of an AP are 4 and –8. Which term is zero?

Solution

$a + 2 d = 4$ , $a + 8 d = - 8 \Rightarrow 6 d = - 12 \Rightarrow d = - 2$ , $a = 8$ .

$0 = 8 + (n - 1) (- 2) \Rightarrow n = 5$

The 5th term.

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution

$(a + 16 d) - (a + 9 d) = 7 \Rightarrow 7 d = 7$

d = 1

Which term of 3, 15, 27, 39, … is 132 more than its 54th term?

Solution

$a_{54} = 3 + 53 \times 12 = 639$ , so we need $639 + 132 = 771$ .

$771 = 3 + (n - 1) 12 \Rightarrow n - 1 = 64 \Rightarrow n = 65$

(Quicker: 132 more is $132 \div 12 = 11$ terms later, $54 + 11 = 65$ .)

The 65th term.

Two APs have the same common difference, and their 100th terms differ by 100. What is the difference between their 1000th terms?

Solution

$(a + 99 d) - (b + 99 d) = a - b = 100$ . Then $(a + 999 d) - (b + 999 d) = a - b$ .

100 (the difference stays the same for every term).

How many three-digit numbers are divisible by 7?

Solution

The AP is 105, 112, …, 994: $994 = 105 + (n - 1) 7 \Rightarrow n - 1 = 127 \Rightarrow n = 128$

128

How many multiples of 4 lie between 10 and 250?

Solution

The AP is 12, 16, …, 248: $248 = 12 + (n - 1) 4 \Rightarrow n = 60$

For what n are the nth terms of 63, 65, 67, … and 3, 10, 17, … equal?

Solution

$63 + 2 (n - 1) = 3 + 7 (n - 1) \Rightarrow 60 = 5 (n - 1) \Rightarrow n = 13$

(Both 13th terms equal 87.)

n = 13

Find the AP whose third term is 16 and whose 7th term exceeds its 5th term by 12.

Solution

$a_{7} - a_{5} = 2 d = 12 \Rightarrow d = 6$ ; $a + 2 d = 16 \Rightarrow a = 4$ .

4, 10, 16, 22, …

Find the 20th term from the last of the AP 3, 8, 13, …, 253.

Solution

Reading backwards gives an AP with $a = 253$ , $d = - 5$ : $a_{20} = 253 + 19 (- 5) = 158$

158

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

Solution

$2 a + 10 d = 24$ and $2 a + 14 d = 44 \Rightarrow 4 d = 20 \Rightarrow d = 5$ , $a = - 13$ .

–13, –8, –3

Subba Rao started work in 1995 at ₹5000 a year with an increment of ₹200 each year. In which year did his income reach ₹7000?

Solution

$7000 = 5000 + (n - 1) 200 \Rightarrow n - 1 = 10 \Rightarrow n = 11$ . The 11th year from 1995 is 2005.

In 2005 (his 11th year).

Ramkali saved ₹5 in the first week and increased her weekly saving by ₹1.75. In the nth week she saved ₹20.75. Find n.

Solution

$20.75 = 5 + (n - 1) 1.75 \Rightarrow n - 1 = 9 \Rightarrow n = 10$

n = 10

Exercise 5.3

Find the sums: (i) 2, 7, 12, … to 10 terms (ii) –37, –33, –29, … to 12 terms (iii) 0.6, 1.7, 2.8, … to 100 terms (iv) 1/15, 1/12, 1/10, … to 11 terms

Solution

$S_{10} = 5 [4 + 9 \times 5] = 5 \times 49 = 245$
$S_{12} = 6 [- 74 + 11 \times 4] = 6 \times (- 30) = - 180$
$S_{100} = 50 [1.2 + 99 \times 1.1] = 50 \times 110.1 = 5505$
$d = \frac{1}{12} - \frac{1}{15} = \frac{1}{60}$ : $S_{11} = \frac{11}{2} [\frac{2}{15} + \frac{10}{60}] = \frac{11}{2} \times \frac{9}{30} = \frac{33}{20}$

(i) 245 (ii) –180 (iii) 5505 (iv) 33/20

Find the sums: (i) 7 + 10½ + 14 + … + 84 (ii) 34 + 32 + 30 + … + 10 (iii) –5 + (–8) + (–11) + … + (–230)

Solution

$d = 3.5$ : $84 = 7 + (n - 1) 3.5 \Rightarrow n = 23$ . $S = \frac{23}{2} (7 + 84) = 1046 \frac{1}{2}$
$d = - 2$ : $10 = 34 - 2 (n - 1) \Rightarrow n = 13$ . $S = \frac{13}{2} (34 + 10) = 286$
$d = - 3$ : $- 230 = - 5 - 3 (n - 1) \Rightarrow n = 76$ . $S = 38 (- 5 - 230) = - 8930$

(i) 1046½ (ii) 286 (iii) –8930

In an AP: (i) a = 5, d = 3, aₙ = 50; find n and Sₙ. (ii) a = 7, a₁₃ = 35; find d and S₁₃. (iii) a₁₂ = 37, d = 3; find a and S₁₂. (iv) a₃ = 15, S₁₀ = 125; find d and a₁₀. (v) d = 5, S₉ = 75; find a and a₉. (vi) a = 2, d = 8, Sₙ = 90; find n and aₙ. (vii) a = 8, aₙ = 62, Sₙ = 210; find n and d. (viii) aₙ = 4, d = 2, Sₙ = –14; find n and a. (ix) a = 3, n = 8, S = 192; find d. (x) l = 28, S = 144, 9 terms; find a.

Solution

$50 = 5 + 3 (n - 1) \Rightarrow n = 16$ ; $S_{16} = 8 (5 + 50) = 440$ .
$35 = 7 + 12 d \Rightarrow d = \frac{7}{3}$ ; $S_{13} = \frac{13}{2} (7 + 35) = 273$ .
$37 = a + 33 \Rightarrow a = 4$ ; $S_{12} = 6 (4 + 37) = 246$ .
$a + 2 d = 15$ and $5 (2 a + 9 d) = 125$ , i.e. $2 a + 9 d = 25$ . Putting $a = 15 - 2 d$ : $30 + 5 d = 25 \Rightarrow d = - 1$ , $a = 17$ ; $a_{10} = 17 - 9 = 8$ .
$\frac{9}{2} (2 a + 40) = 75 \Rightarrow 2 a = \frac{50}{3} - 40 \Rightarrow a = - \frac{35}{3}$ ; $a_{9} = - \frac{35}{3} + 40 = \frac{85}{3}$ .
$\frac{n}{2} [4 + 8 (n - 1)] = 90 \Rightarrow 2 n^{2} - n - 45 = 0 \Rightarrow (2 n + 9) (n - 5) = 0 \Rightarrow n = 5$ ; $a_{5} = 2 + 32 = 34$ .
$\frac{n}{2} (8 + 62) = 210 \Rightarrow n = 6$ ; $62 = 8 + 5 d \Rightarrow d = \frac{54}{5}$ .
$a = 4 - 2 (n - 1) = 6 - 2 n$ ; $\frac{n}{2} (a + 4) = - 14 \Rightarrow n (5 - n) = - 14 \Rightarrow n^{2} - 5 n - 14 = 0 \Rightarrow n = 7$ ; $a = - 8$ .
$4 (6 + 7 d) = 192 \Rightarrow 7 d = 42 \Rightarrow d = 6$ .
$\frac{9}{2} (a + 28) = 144 \Rightarrow a + 28 = 32 \Rightarrow a = 4$ .

(i) n = 16, S = 440 (ii) d = 7/3, S = 273 (iii) a = 4, S = 246 (iv) d = –1, a₁₀ = 8 (v) a = –35/3, a₉ = 85/3 (vi) n = 5, aₙ = 34 (vii) n = 6, d = 54/5 (viii) n = 7, a = –8 (ix) d = 6 (x) a = 4

How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?

Solution

$\frac{n}{2} [18 + 8 (n - 1)] = 636 \Rightarrow 4 n^{2} + 5 n - 636 = 0 \Rightarrow (n - 12) (4 n + 53) = 0$

$n$ must be a positive whole number, so $n = 12$ .

12 terms

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

Solution

$\frac{n}{2} (5 + 45) = 400 \Rightarrow n = 16$ ; $45 = 5 + 15 d \Rightarrow d = \frac{8}{3}$

16 terms, d = 8/3

The first and last terms of an AP are 17 and 350, and d = 9. How many terms are there, and what is their sum?

Solution

$350 = 17 + 9 (n - 1) \Rightarrow n = 38$ ; $S = 19 (17 + 350) = 6973$

38 terms, sum 6973

Find the sum of the first 22 terms of an AP with d = 7 and 22nd term 149.

Solution

$149 = a + 21 \times 7 \Rightarrow a = 2$ ; $S_{22} = 11 (2 + 149) = 1661$

1661

Find the sum of the first 51 terms of an AP whose 2nd and 3rd terms are 14 and 18.

Solution

$d = 4$ , $a = 10$ : $S_{51} = \frac{51}{2} [20 + 50 \times 4] = 51 \times 110 = 5610$

5610

The sum of the first 7 terms of an AP is 49 and of the first 17 terms is 289. Find the sum of the first n terms.

Solution

$\frac{7}{2} (2 a + 6 d) = 49 \Rightarrow a + 3 d = 7$ ; $\frac{17}{2} (2 a + 16 d) = 289 \Rightarrow a + 8 d = 17$ . So $d = 2$ , $a = 1$ .

$S_{n} = \frac{n}{2} [2 + 2 (n - 1)] = n^{2}$

Sₙ = n²

Show that the sequences are APs and find the sum of the first 15 terms: (i) aₙ = 3 + 4n (ii) aₙ = 9 – 5n

Solution

$a_{n + 1} - a_{n}$ is a constant in each case, so each is an AP.

$a_{n + 1} - a_{n} = 4$ . $a_{1} = 7$ , $a_{15} = 63$ , $S_{15} = \frac{15}{2} (7 + 63) = 525$ .
$a_{n + 1} - a_{n} = - 5$ . $a_{1} = 4$ , $a_{15} = - 66$ , $S_{15} = \frac{15}{2} (4 - 66) = - 465$ .

(i) d = 4, S₁₅ = 525 (ii) d = –5, S₁₅ = –465

If the sum of the first n terms of an AP is 4n – n², find S₁, S₂, the second term, the 3rd, 10th and nth terms.

Solution

$S_{1} = 3$ , so $a_{1} = 3$ . $S_{2} = 4$ , so $a_{2} = S_{2} - S_{1} = 1$ . $S_{3} = 3$ , so $a_{3} = 3 - 4 = - 1$ .

In general, $a_{n} = S_{n} - S_{n - 1} = (4 n - n^{2}) - [4 (n - 1) - (n - 1)^{2}] = 5 - 2 n$ , so $a_{10} = - 15$ .

S₁ = 3, S₂ = 4, a₂ = 1, a₃ = –1, a₁₀ = –15, aₙ = 5 – 2n

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

Solution

6, 12, …, 240: $S_{40} = 20 (6 + 240) = 4920$

4920

Find the sum of the first 15 multiples of 8.

Solution

8, 16, …, 120: $S_{15} = \frac{15}{2} (8 + 120) = 960$

960

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

Solution

1, 3, …, 49 (25 terms): $S = \frac{25}{2} (1 + 49) = 625$

625

A delay penalty is ₹200 for the first day, ₹250 for the second, ₹300 for the third, and so on. How much is the penalty for a 30-day delay?

Solution

$a = 200$ , $d = 50$ : $S_{30} = 15 [400 + 29 \times 50] = 15 \times 1850 = 27750$

₹27,750

₹700 is given as seven prizes, each ₹20 less than the one before. Find each prize.

Solution

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

₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

Each section of Class I plants 1 tree, of Class II 2 trees, …, of Class XII 12 trees. Each class has three sections. How many trees are planted?

Solution

Each class plants $3 \times$ its class number: $3 (1 + 2 + \dots + 12) = 3 \times 78 = 234$

234 trees

A spiral is made of 13 semicircles with centres alternately at A and B and radii 0.5 cm, 1.0 cm, 1.5 cm, …. Find its total length (π = 22/7).

Solution

Each semicircle has length $π r$ , so the lengths are $π (0.5), π (1.0), \dots, π (6.5)$ : an AP.

Total $= π \times \frac{13}{2} (0.5 + 6.5) = \frac{22}{7} \times 45.5 = 143$

143 cm

200 logs are stacked with 20 in the bottom row, 19 in the next, 18 in the next and so on. How many rows are there, and how many logs are in the top row?

Solution

$\frac{n}{2} [40 - (n - 1)] = 200 \Rightarrow n^{2} - 41 n + 400 = 0 \Rightarrow (n - 16) (n - 25) = 0$

$n = 25$ would give $a_{25} = 20 - 24 = - 4$ logs, which is impossible. So $n = 16$ and the top row has $20 - 15 = 5$ logs.

16 rows, with 5 logs in the top row.

In a potato race, the bucket is 5 m from the first potato, the other potatoes are 3 m apart, and there are ten potatoes. Running to each potato and back to the bucket, what total distance does the competitor run?

Solution

The distances to the potatoes are 5, 8, 11, …, 32 m, and each is run twice:

$2 \times \frac{10}{2} (5 + 32) = 370$

370 m

Exercise 5.4 (Optional)

Which term of the AP 121, 117, 113, … is its first negative term?

Solution

$a_{n} = 121 - 4 (n - 1) < 0 \Rightarrow n - 1 > 30.25 \Rightarrow n \geq 32$ . ( $a_{31} = 1$ , $a_{32} = - 3$ .)

The 32nd term.

The sum of the 3rd and 7th terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms.

Solution

$2 a + 8 d = 6 \Rightarrow a + 4 d = 3$ . So the terms are $3 - 2 d$ and $3 + 2 d$ , and $(3 - 2 d) (3 + 2 d) = 8 \Rightarrow d^{2} = \frac{1}{4} \Rightarrow d = \pm \frac{1}{2}$ .

$d = \frac{1}{2}$ , $a = 1$ : $S_{16} = 8 [2 + 15 \times \frac{1}{2}] = 76$
$d = - \frac{1}{2}$ , $a = 5$ : $S_{16} = 8 [10 - \frac{15}{2}] = 20$

76 (if d = 1/2) or 20 (if d = –1/2)

A ladder has rungs 25 cm apart, decreasing uniformly from 45 cm at the bottom to 25 cm at the top. The top and bottom rungs are 2½ m apart. What length of wood is needed for the rungs?

Solution

Number of rungs $= \frac{250}{25} + 1 = 11$ . Total length $= \frac{11}{2} (45 + 25) = 385$ .

385 cm

Houses are numbered 1 to 49. Show that there is an x such that the sum of the numbers before house x equals the sum of the numbers after it, and find x.

Solution

$S_{x - 1} = S_{49} - S_{x} \Rightarrow \frac{x ( x - 1 )}{2} = 1225 - \frac{x ( x + 1 )}{2} \Rightarrow x^{2} = 1225 \Rightarrow x = 35$

Check: $1 + \dots + 34 = 595$ and $36 + \dots + 49 = 1225 - 630 = 595$ ✓

x = 35

A terrace has 15 steps, each 50 m long, with a rise of ¼ m and a tread of ½ m. Find the total volume of concrete.

Solution

The $k$ th step is a solid block $50 \times \frac{1}{2} \times \frac{k}{4}$ m³ $= \frac{25 k}{4}$ m³.

Total $= \frac{25}{4} (1 + 2 + \dots + 15) = \frac{25}{4} \times 120 = 750$

750 m³

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