Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 Solutions | English Medium | SEBA/SCERT Assam 2026–2027

Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 Solutions | English Medium | SEBA/SCERT Assam 2026–2027

 

Are you looking for the best Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 Solutions in English Medium? Here you will get complete and step-by-step solutions based on the latest SEBA/SCERT Assam 2026–2027 syllabus.

Exercise 5.3 is one of the most important exercises of Arithmetic Progression (AP). In this exercise, students learn how to find the sum of n terms of an AP using formulas and shortcuts.

Exercise 5.3 – Topics Covered

In Exercise 5.3, students learn:

Sum of n terms of AP
Formula application
Finding total value of sequences
AP-based word problems
Large number calculations

Get complete Class 10 Maths Chapter 5 Exercise 5.3 solutions in English Medium based on SEBA/SCERT Assam 2026–2027 syllabus. Learn Arithmetic Progressions sum formulas and step-by-step solutions easily.

 

Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3 Solutions | English Medium (SEBA/SCERT Assam 2026–2027)

Exercise 5.3

1. Find the sum of the following APs:

(i) \(2,\;7,\;12,\;\dots\) (to 10 terms)

Solution:
Here, \(a = 2\)
\(d = 7 - 2 = 5\)
\(n = 10\)
We know that,
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_{10} = \frac{10}{2}[2(2) + (10-1)5]\)
\(\Rightarrow S_{10} = 5[4 + 9 \times 5]\)
\(\Rightarrow S_{10} = 5[4 + 45]\)
\(\Rightarrow S_{10} = 5 \times 49\)
\(\Rightarrow S_{10} = 245\)
\(\therefore\) The required sum is 245.

(ii) \(-37,\;-33,\;-29,\;\dots\) (to 12 terms)

Solution:
Here, \(a = -37\)
\(d = -33 - (-37) = 4\)
\(n = 12\)
\(\Rightarrow S_{12} = \frac{12}{2}[2(-37) + (12-1)4]\)
\(\Rightarrow S_{12} = 6[-74 + 11 \times 4]\)
\(\Rightarrow S_{12} = 6[-74 + 44]\)
\(\Rightarrow S_{12} = 6[-30]\)
\(\Rightarrow S_{12} = -180\)
\(\therefore\) The required sum is -180.

(iii) \(0.6,\;1.7,\;2.8,\;\dots\) (to 100 terms)

Solution:
Here, \(a = 0.6, d = 1.1, n = 100\)
\(\Rightarrow S_{100} = \frac{100}{2}[2(0.6) + (100-1)1.1]\)
\(\Rightarrow S_{100} = 50[1.2 + 99 \times 1.1]\)
\(\Rightarrow S_{100} = 50[1.2 + 108.9]\)
\(\Rightarrow S_{100} = 50[110.1]\)
\(\Rightarrow S_{100} = 5505\)
\(\therefore\) The required sum is 5505.

(iv) \(\frac{1}{15},\;\frac{1}{12},\;\frac{1}{10},\;\dots\) (to 11 terms)

Solution:
Here, \(a = \frac{1}{15}, n = 11\)
\(d = \frac{1}{12} - \frac{1}{15} = \frac{5-4}{60} = \frac{1}{60}\)
\(\Rightarrow S_{11} = \frac{11}{2}[2(\frac{1}{15}) + (11-1)\frac{1}{60}] \)
\(\Rightarrow S_{11} = \frac{11}{2}[\frac{2}{15} + 10 \times \frac{1}{60}] \)
\(\Rightarrow S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{1}{6}] \)
\(\Rightarrow S_{11} = \frac{11}{2}[\frac{4+5}{30}] \)
\(\Rightarrow S_{11} = \frac{11}{2} \times \frac{9}{30} \)
\(\Rightarrow S_{11} = \frac{33}{20}\) or \(1.65\)
\(\therefore\) The required sum is \(\frac{33}{20}\).

2. Find the sums given below:

(i) \(7+10\frac{1}{2}+14+\dots+84\)

Solution:
Here, \(a = 7, d = 3.5 = \frac{7}{2}, a_n = 84\)
\(\Rightarrow a + (n-1)d = 84\)
\(\Rightarrow 7 + (n-1)\frac{7}{2} = 84\)
\(\Rightarrow (n-1)\frac{7}{2} = 77\)
\(\Rightarrow n-1 = 77 \times \frac{2}{7}\)
\(\Rightarrow n-1 = 22 \Rightarrow n = 23\)
\(\Rightarrow S_{23} = \frac{23}{2}(7 + 84)\)
\(\Rightarrow S_{23} = \frac{23}{2} \times 91\)
\(\Rightarrow S_{23} = \frac{2093}{2} = 1046.5\)
\(\therefore\) The required sum is 1046.5.

(ii) \(34+32+30+\dots+10\)

Solution:
Here, \(a = 34, d = -2, a_n = 10\)
\(\Rightarrow 34 + (n-1)(-2) = 10\)
\(\Rightarrow (n-1)(-2) = -24\)
\(\Rightarrow n-1 = 12 \Rightarrow n = 13\)
\(\Rightarrow S_{13} = \frac{13}{2}(34 + 10)\)
\(\Rightarrow S_{13} = \frac{13}{2} \times 44\)
\(\Rightarrow S_{13} = 13 \times 22\)
\(\Rightarrow S_{13} = 286\)
\(\therefore\) The required sum is 286.

(iii) \(-5+(-8)+(-11)+\dots+(-230)\)

Solution:
Here, \(a = -5, d = -3, a_n = -230\)
\(\Rightarrow -5 + (n-1)(-3) = -230\)
\(\Rightarrow (n-1)(-3) = -225\)
\(\Rightarrow n-1 = 75 \Rightarrow n = 76\)
\(\Rightarrow S_{76} = \frac{76}{2}[-5 + (-230)]\)
\(\Rightarrow S_{76} = 38 \times (-235)\)
\(\Rightarrow S_{76} = -8930\)
\(\therefore\) The required sum is -8930.

3. In an AP—

(i) Given \(a=5,\; d=3,\; a_n=50\), find \(n\) and \(S_n\).

Solution:
Here, \(a=5, d=3, a_n=50\)
We know that, \(a_n = a + (n-1)d\)
\(\Rightarrow 50 = 5 + (n-1)3\)
\(\Rightarrow 50 - 5 = 3(n-1)\)
\(\Rightarrow 45 = 3(n-1)\)
\(\Rightarrow n-1 = 15\)
\(\Rightarrow n = 16\)
Now, \(S_n = \frac{n}{2}(a + a_n)\)
\(\Rightarrow S_{16} = \frac{16}{2}(5 + 50)\)
\(\Rightarrow S_{16} = 8 \times 55\)
\(\Rightarrow S_{16} = 440\)
\(\therefore n = 16\) and \(S_n = 440\).

(ii) Given \(a=7,\; a_{13}=35\), find \(d\) and \(S_{13}\).

Solution:
Here, \(a=7, a_{13}=35, n=13\)
\(\Rightarrow a + (13-1)d = 35\)
\(\Rightarrow 7 + 12d = 35\)
\(\Rightarrow 12d = 35 - 7\)
\(\Rightarrow 12d = 28\)
\(\Rightarrow d = \frac{28}{12} = \frac{7}{3}\)
Now, \(S_{13} = \frac{13}{2}(a + a_{13})\)
\(\Rightarrow S_{13} = \frac{13}{2}(7 + 35)\)
\(\Rightarrow S_{13} = \frac{13}{2} \times 42\)
\(\Rightarrow S_{13} = 13 \times 21 = 273\)
\(\therefore d = \frac{7}{3}\) and \(S_{13} = 273\).

(iii) Given \(a_{12}=37,\; d=3\), find \(a\) and \(S_{12}\).

Solution:
Here, \(a_{12}=37, d=3, n=12\)
\(\Rightarrow a + (12-1)3 = 37\)
\(\Rightarrow a + 11 \times 3 = 37\)
\(\Rightarrow a + 33 = 37\)
\(\Rightarrow a = 4\)
Now, \(S_{12} = \frac{12}{2}(a + a_{12})\)
\(\Rightarrow S_{12} = 6(4 + 37)\)
\(\Rightarrow S_{12} = 6 \times 41 = 246\)
\(\therefore a = 4\) and \(S_{12} = 246\).

(iv) Given \(a_3=15,\; S_{10}=125\), find \(d\) and \(a_{10}\).

Solution:
\(a_3 = 15 \Rightarrow a + 2d = 15\) ----(1)
\(S_{10} = 125 \Rightarrow \frac{10}{2}[2a + (10-1)d] = 125\)
\(\Rightarrow 5[2a + 9d] = 125\)
\(\Rightarrow 2a + 9d = 25\) ----(2)
Multiplying eq (1) by 2 and subtracting from (2):
\(\Rightarrow (2a + 9d) - (2a + 4d) = 25 - 30\)
\(\Rightarrow 5d = -5\)
\(\Rightarrow d = -1\)
Now from (1), \(a + 2(-1) = 15\)
\(\Rightarrow a = 17\)
Now, \(a_{10} = a + 9d\)
\(\Rightarrow a_{10} = 17 + 9(-1) = 8\)
\(\therefore d = -1\) and \(a_{10} = 8\).

(v) Given \(d=5,\; S_9=75\), find \(a\) and \(a_9\).

Solution:
\(S_9 = 75 \Rightarrow \frac{9}{2}[2a + (9-1)5] = 75\)
\(\Rightarrow \frac{9}{2}[2a + 40] = 75\)
\(\Rightarrow 9(a + 20) = 75\)
\(\Rightarrow 9a + 180 = 75\)
\(\Rightarrow 9a = -105\)
\(\Rightarrow a = -\frac{35}{3}\)
Now, \(a_9 = a + 8d\)
\(\Rightarrow a_9 = -\frac{35}{3} + 8(5)\)
\(\Rightarrow a_9 = -\frac{35}{3} + 40\)
\(\Rightarrow a_9 = \frac{-35 + 120}{3}\)
\(\Rightarrow a_9 = \frac{85}{3}\)
\(\therefore a = -\frac{35}{3}\) and \(a_9 = \frac{85}{3}\).

(vi) Given \(a=2,\; d=8,\; S_n=90\), find \(n\) and \(a_n\).

Solution:
\(S_n = 90 \Rightarrow \frac{n}{2}[2(2) + (n-1)8] = 90\)
\(\Rightarrow \frac{n}{2}[4 + 8n - 8] = 90\)
\(\Rightarrow \frac{n}{2}[8n - 4] = 90\)
\(\Rightarrow n(4n - 2) = 90\)
\(\Rightarrow 4n^2 - 2n - 90 = 0\)
\(\Rightarrow 2n^2 - n - 45 = 0\)
\(\Rightarrow 2n^2 - 10n + 9n - 45 = 0\)
\(\Rightarrow 2n(n - 5) + 9(n - 5) = 0\)
\(\Rightarrow (n-5)(2n+9) = 0\)
\(\Rightarrow n = 5\) (Since \(n > 0\))
Now, \(a_5 = a + 4d\)
\(\Rightarrow a_5 = 2 + 4(8)\)
\(\Rightarrow a_5 = 34\)
\(\therefore n = 5\) and \(a_n = 34\).

(vii) Given \(a=8,\; a_n=62,\; S_n=210\), find \(n\) and \(d\).

Solution:
\(S_n = \frac{n}{2}(a + a_n)\)
\(\Rightarrow 210 = \frac{n}{2}(8 + 62)\)
\(\Rightarrow 210 = \frac{n}{2} \times 70\)
\(\Rightarrow 210 = 35n\)
\(\Rightarrow n = 6\)
Now, \(a_n = 62\)
\(\Rightarrow 8 + (6-1)d = 62\)
\(\Rightarrow 5d = 54\)
\(\Rightarrow d = \frac{54}{5} = 10.8\)
\(\therefore n = 6\) and \(d = 10.8\).

(viii) Given \(a_n=4,\; d=2,\; S_n=-14\), find \(n\) and \(a\).

Solution:
\(a_n = 4 \Rightarrow a + (n-1)2 = 4\)
\(\Rightarrow a = 4 - 2n + 2\)
\(\Rightarrow a = 6 - 2n\) ----(1)
Now, \(S_n = -14 \Rightarrow \frac{n}{2}(a + a_n) = -14\)
\(\Rightarrow \frac{n}{2}(6 - 2n + 4) = -14\)
\(\Rightarrow \frac{n}{2}(10 - 2n) = -14\)
\(\Rightarrow n(5 - n) = -14\)
\(\Rightarrow 5n - n^2 = -14\)
\(\Rightarrow n^2 - 5n - 14 = 0\)
\(\Rightarrow (n-7)(n+2) = 0\)
\(\Rightarrow n = 7\) (Since \(n > 0\))
\(\Rightarrow a = 6 - 2(7) = -8\)
\(\therefore n = 7\) and \(a = -8\).

(ix) Given \(a=3,\; n=8,\; S=192\), find \(d\).

Solution:
\(S = 192, n = 8, a = 3\)
\(\Rightarrow \frac{8}{2}[2(3) + (8-1)d] = 192\)
\(\Rightarrow 4[6 + 7d] = 192\)
\(\Rightarrow 6 + 7d = 48\)
\(\Rightarrow 7d = 42\)
\(\Rightarrow d = 6\)
\(\therefore d = 6\).

(x) Given \(l=28,\; S=144\) and total terms \(n=9\), find \(a\).

Solution:
Here, \(l = a_n = 28, S = 144, n = 9\)
\(\Rightarrow S = \frac{n}{2}(a + l)\)
\(\Rightarrow 144 = \frac{9}{2}(a + 28)\)
\(\Rightarrow 144 \times \frac{2}{9} = a + 28\)
\(\Rightarrow 16 \times 2 = a + 28\)
\(\Rightarrow 32 = a + 28\)
\(\Rightarrow a = 4\)
\(\therefore a = 4\).

4. How many terms of the AP \(9,\;17,\;25,\;\dots\) must be taken to give a sum of \(636\)?

Solution:
Here, \(a = 9, d = 17 - 9 = 8, S_n = 636\)
\(\Rightarrow \frac{n}{2}[2(9) + (n-1)8] = 636\)
\(\Rightarrow \frac{n}{2}[18 + 8n - 8] = 636\)
\(\Rightarrow \frac{n}{2}[10 + 8n] = 636\)
\(\Rightarrow n(5 + 4n) = 636\)
\(\Rightarrow 4n^2 + 5n - 636 = 0\)
\(\Rightarrow 4n^2 + 53n - 48n - 636 = 0\)
\(\Rightarrow n(4n + 53) - 12(4n + 53) = 0\)
\(\Rightarrow (n - 12)(4n + 53) = 0\)
\(\Rightarrow n = 12\) (Since \(n > 0\))
\(\therefore\) 12 terms must be taken to get a sum of 636.

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:
Here, \(a = 5, l = 45, S_n = 400\)
\(\Rightarrow \frac{n}{2}(a + l) = 400\)
\(\Rightarrow \frac{n}{2}(5 + 45) = 400\)
\(\Rightarrow 25n = 400 \Rightarrow n = 16\)
Now, \(l = a + (n-1)d\)
\(\Rightarrow 45 = 5 + (16-1)d\)
\(\Rightarrow 40 = 15d \Rightarrow d = \frac{40}{15} = \frac{8}{3}\)
\(\therefore\) Number of terms is 16 and common difference is \(\frac{8}{3}\).

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:
Given,
First term, \(a = 17\)
Last term, \(l = a_n = 350\)
Common difference, \(d = 9\)
We know that,
\(a_n = a + (n-1)d\)
\(\Rightarrow 350 = 17 + (n-1)9\)
\(\Rightarrow 350 - 17 = 9(n-1)\)
\(\Rightarrow 333 = 9(n-1)\)
\(\Rightarrow n-1 = \frac{333}{9}\)
\(\Rightarrow n-1 = 37\)
\(\Rightarrow n = 37 + 1\)
\(\Rightarrow n = 38\)
Now, for the sum:
\(S_n = \frac{n}{2}(a + a_n)\)
\(\Rightarrow S_{38} = \frac{38}{2}(17 + 350)\)
\(\Rightarrow S_{38} = 19(367)\)
\(\Rightarrow S_{38} = 6973\)
\(\therefore\) There are 38 terms and their sum is 6973.

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

Solution:
Given,
Common difference, \(d = 7\)
\(22\)nd term, \(a_{22} = 149\)
Number of terms, \(n = 22\)
We know that,
\(a_n = a + (n-1)d\)
\(\Rightarrow 149 = a + (22-1)7\)
\(\Rightarrow 149 = a + 21 \times 7\)
\(\Rightarrow 149 = a + 147\)
\(\Rightarrow a = 149 - 147\)
\(\Rightarrow a = 2\)
Now, sum of first \(22\) terms:
\(S_n = \frac{n}{2}(a + a_n)\)
\(\Rightarrow S_{22} = \frac{22}{2}(2 + 149)\)
\(\Rightarrow S_{22} = 11(151)\)
\(\Rightarrow S_{22} = 1661\)
\(\therefore\) The required sum is 1661.

 

8. Find the sum of the first \(51\) terms of an AP whose second and third terms are \(14\) and \(18\) respectively.

Solution:
Given that,
Second term, \(a_2 = 14\)
Third term, \(a_3 = 18\)
\(\therefore\) Common difference, \(d = a_3 - a_2\)
\(\Rightarrow d = 18 - 14\)
\(\Rightarrow d = 4\)
We know that, \(a_2 = a + d\)
\(\Rightarrow 14 = a + 4\)
\(\Rightarrow a = 14 - 4\)
\(\Rightarrow a = 10\)
Now, the sum of the first \(51\) terms (\(n = 51\)),
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_{51} = \frac{51}{2}[2(10) + (51-1)4]\)
\(\Rightarrow S_{51} = \frac{51}{2}[20 + 50 \times 4]\)
\(\Rightarrow S_{51} = \frac{51}{2}[20 + 200]\)
\(\Rightarrow S_{51} = \frac{51}{2} \times 220\)
\(\Rightarrow S_{51} = 51 \times 110\)
\(\Rightarrow S_{51} = 5610\)
\(\dots\) The sum of the first 51 terms is 5610.

9. If the sum of the first \(7\) terms of an AP is \(49\) and that of \(17\) terms is \(289\), find the sum of the first \(n\) terms.

Solution:
Given that, \(S_7 = 49\) and \(S_{17} = 289\)
We know that, \(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_7 = \frac{7}{2}[2a + (7-1)d] = 49\)
\(\Rightarrow \frac{7}{2}[2a + 6d] = 49\)
\(\Rightarrow 7[a + 3d] = 49\)
\(\Rightarrow a + 3d = 7\) ----(1)
Similarly,
\(\Rightarrow S_{17} = \frac{17}{2}[2a + (17-1)d] = 289\)
\(\Rightarrow \frac{17}{2}[2a + 16d] = 289\)
\(\Rightarrow 17[a + 8d] = 289\)
\(\Rightarrow a + 8d = \frac{289}{17}\)
\(\Rightarrow a + 8d = 17\) ----(2)
Subtracting equation (1) from (2), we get:
\(\Rightarrow (a + 8d) - (a + 3d) = 17 - 7\)
\(\Rightarrow 5d = 10\)
\(\Rightarrow d = 2\)
Now, substituting \(d = 2\) in equation (1), we get:
\(\Rightarrow a + 3(2) = 7\)
\(\Rightarrow a + 6 = 7\)
\(\Rightarrow a = 1\)
Now, the sum of the first \(n\) terms,
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_n = \frac{n}{2}[2(1) + (n-1)2]\)
\(\Rightarrow S_n = \frac{n}{2}[2 + 2n - 2]\)
\(\Rightarrow S_n = \frac{n}{2}[2n]\)
\(\Rightarrow S_n = n^2\)
\(\dots\) The sum of the first \(n\) terms is \(n^2\).

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.

(i) \(a_n = 3 + 4n\)

Solution:
Here, \(a_n = 3 + 4n\)
Substituting \(n = 1, 2, 3, \dots\), we get:
\(\Rightarrow a_1 = 3 + 4(1) = 7\)
\(\Rightarrow a_2 = 3 + 4(2) = 11\)
\(\Rightarrow a_3 = 3 + 4(3) = 15\)
Here the terms are: \(7, 11, 15, \dots\)
Now, let's check the differences:
\(\Rightarrow a_2 - a_1 = 11 - 7 = 4\)
\(\Rightarrow a_3 - a_2 = 15 - 11 = 4\)
Since the common difference \(d = 4\) (always constant), it forms an AP.
Here, first term \(a = 7\), common difference \(d = 4\) and number of terms \(n = 15\).
\(\therefore S_{15} = \frac{15}{2}[2(7) + (15-1)4]\)
\(\Rightarrow S_{15} = \frac{15}{2}[14 + 14 \times 4]\)
\(\Rightarrow S_{15} = \frac{15}{2}[14 + 56]\)
\(\Rightarrow S_{15} = \frac{15}{2} \times 70\)
\(\Rightarrow S_{15} = 15 \times 35\)
\(\Rightarrow S_{15} = 525\)
\(\dots\) The sum of the first 15 terms is 525.

(ii) \(a_n = 9 - 5n\)

Solution:
Here, \(a_n = 9 - 5n\)
Substituting \(n = 1, 2, 3, \dots\), we get:
\(\Rightarrow a_1 = 9 - 5(1) = 4\)
\(\Rightarrow a_2 = 9 - 5(2) = -1\)
\(\Rightarrow a_3 = 9 - 5(3) = -6\)
Here the terms are: \(4, -1, -6, \dots\)
Now, let's check the differences:
\(\Rightarrow a_2 - a_1 = -1 - 4 = -5\)
\(\Rightarrow a_3 - a_2 = -6 - (-1) = -5\)
Since the common difference \(d = -5\) (always constant), it forms an AP.
Here, first term \(a = 4\), common difference \(d = -5\) and number of terms \(n = 15\).
\(\therefore S_{15} = \frac{15}{2}[2(4) + (15-1)(-5)]\)
\(\Rightarrow S_{15} = \frac{15}{2}[8 + 14 \times (-5)]\)
\(\Rightarrow S_{15} = \frac{15}{2}[8 - 70]\)
\(\Rightarrow S_{15} = \frac{15}{2} \times (-62)\)
\(\Rightarrow S_{15} = 15 \times (-31)\)
\(\Rightarrow S_{15} = -465\)
\(\dots\) The sum of the first 15 terms is -465.

11. If the sum of the first \(n\) terms of an AP is \(S_n = 4n - n^2\), then 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 third, the tenth and the \(n\)-th terms.

Solution:
Given, \(S_n = 4n - n^2\)
Substituting \(n = 1\), we get the first term:
\(\Rightarrow S_1 = 4(1) - (1)^2\)
\(\Rightarrow S_1 = 4 - 1 = 3\)
\(\therefore\) First term \(a_1 = S_1 = 3\)

Substituting \(n = 2\), we get the sum of the first two terms:
\(\Rightarrow S_2 = 4(2) - (2)^2\)
\(\Rightarrow S_2 = 8 - 4 = 4\)
\(\dots\) The sum of the first two terms is 4.

\(\therefore\) Second term, \(a_2 = S_2 - S_1\)
\(\Rightarrow a_2 = 4 - 3 = 1\)
\(\dots\) Common difference, \(d = a_2 - a_1\)
\(\Rightarrow d = 1 - 3 = -2\)

Now let's find the other terms:
Third term, \(a_3 = a + 2d\)
\(\Rightarrow a_3 = 3 + 2(-2) = 3 - 4 = -1\)
Tenth term, \(a_{10} = a + 9d\)
\(\Rightarrow a_{10} = 3 + 9(-2) = 3 - 18 = -15\)
\(n\)-th term, \(a_n = a + (n-1)d\)
\(\Rightarrow a_n = 3 + (n-1)(-2)\)
\(\Rightarrow a_n = 3 - 2n + 2\)
\(\Rightarrow a_n = 5 - 2n\)
\(\dots S_1 = 3,\; S_2 = 4,\; a_2 = 1,\; a_3 = -1,\; a_{10} = -15,\; a_n = 5 - 2n\).

12. Find the sum of the first \(40\) positive integers divisible by \(6\).

Solution:
The AP of the first \(40\) positive integers divisible by \(6\) is:
\(6, 12, 18, 24, \dots\)
Here, first term \(a = 6\)
Common difference \(d = 6\)
Number of terms \(n = 40\)
We know that, \(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_{40} = \frac{40}{2}[2(6) + (40-1)6]\)
\(\Rightarrow S_{40} = 20[12 + 39 \times 6]\)
\(\Rightarrow S_{40} = 20[12 + 234]\)
\(\Rightarrow S_{40} = 20 \times 246\)
\(\Rightarrow S_{40} = 4920\)
\(\dots\) The required sum is 4920.

13. Find the sum of the first \(15\) multiples of \(8\).

Solution:
The AP of the multiples of \(8\) is:
\(8, 16, 24, 32, \dots\)
Here, first term \(a = 8\)
Common difference \(d = 8\)
Number of terms \(n = 15\)
We know that, \(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_{15} = \frac{15}{2}[2(8) + (15-1)8]\)
\(\Rightarrow S_{15} = \frac{15}{2}[16 + 14 \times 8]\)
\(\Rightarrow S_{15} = \frac{15}{2}[16 + 112]\)
\(\Rightarrow S_{15} = \frac{15}{2} \times 128\)
\(\Rightarrow S_{15} = 15 \times 64\)
\(\Rightarrow S_{15} = 960\)
\(\dots\) The sum of the first 15 multiples is 960.

14. Find the sum of the odd numbers between \(0\) and \(50\).

Solution:
The AP of the odd numbers between \(0\) and \(50\) is:
\(1, 3, 5, 7, \dots, 49\)
Here, first term \(a = 1\)
Common difference \(d = 3 - 1 = 2\)
Last term \(a_n = 49\)
First, let's find the number of terms (\(n\)):
\(a_n = a + (n-1)d\)
\(\Rightarrow 49 = 1 + (n-1)2\)
\(\Rightarrow 49 - 1 = 2(n-1)\)
\(\Rightarrow 48 = 2(n-1)\)
\(\Rightarrow n-1 = 24\)
\(\Rightarrow n = 25\)
Now, let's find the sum:
\(S_n = \frac{n}{2}(a + a_n)\)
\(\Rightarrow S_{25} = \frac{25}{2}(1 + 49)\)
\(\Rightarrow S_{25} = \frac{25}{2} \times 50\)
\(\Rightarrow S_{25} = 25 \times 25\)
\(\Rightarrow S_{25} = 625\)
\(\dots\) The required sum is 625.

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:
According to the question, the penalty amounts form an Arithmetic Progression (AP):
\(200, 250, 300, 350, \dots\)
Here, first term \(a = 200\)
Common difference \(d = 50\)
Total days (number of terms) \(n = 30\)
The total penalty to be paid by the contractor is the sum of 30 terms (\(S_{30}\)):
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_{30} = \frac{30}{2}[2(200) + (30-1)50]\)
\(\Rightarrow S_{30} = 15[400 + 29 \times 50]\)
\(\Rightarrow S_{30} = 15[400 + 1450]\)
\(\Rightarrow S_{30} = 15 \times 1850\)
\(\Rightarrow S_{30} = 27750\)
\(\dots\) The contractor has to pay a total penalty of ₹27,750.

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:
Here, total number of prizes \(n = 7\)
Sum of total money \(S_7 = 700\)
Since each prize is ₹20 less than the preceding one, the common difference \(d = -20\)
We know that, \(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow 700 = \frac{7}{2}[2a + (7-1)(-20)]\)
\(\Rightarrow 700 \times \frac{2}{7} = 2a + 6 \times (-20)\)
\(\Rightarrow 200 = 2a - 120\)
\(\Rightarrow 2a = 200 + 120\)
\(\Rightarrow 2a = 320\)
\(\Rightarrow a = 160\)
\(\dots\) The value of the first prize is ₹160.
Therefore, the values of the prizes in ₹ are respectively:
\(160\text{},\; 140\text{},\; 120\text{},\; 100\text{},\; 80\text{},\; 60\text{},\; 40\text{}\).

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:
Since there are 3 sections in each class, the total number of trees planted by each class is:
Class I \(= 1 \times 3 = 3\) trees
Class II \(= 2 \times 3 = 6\) trees
Class III \(= 3 \times 3 = 9\) trees
\(\dots\)
Class XII \(= 12 \times 3 = 36\) trees
So, it forms an Arithmetic Progression (AP): \(3, 6, 9, \dots, 36\)
Here, first term \(a = 3\)
Common difference \(d = 6 - 3 = 3\)
Number of terms (total classes) \(n = 12\)
Last term \(l = 36\)
\(\therefore\) Total number of trees planted (\(S_{12}\)):
\(S_n = \frac{n}{2}(a + l)\)
\(\Rightarrow S_{12} = \frac{12}{2}(3 + 36)\)
\(\Rightarrow S_{12} = 6 \times 39\)
\(\Rightarrow S_{12} = 234\)
\(\dots\) The students will plant a total of 234 trees.

 

 

18. A spiral is made up of successive semicircles, with centres alternately at \(A\) and \(B\), starting with centre at \(A\), of radii \(0.5\text{ cm}\), \(1.0\text{ cm}\), \(1.5\text{ cm}\), \(2.0\text{ cm}\), \(\ldots\) as shown in Fig. 5.4. What is the total length of such a spiral made up of \(13\) consecutive semicircles? \(\left(\text{Take } \pi=\frac{22}{7}\right)\)

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

Solution:
We know that, circumference (length) of a semicircle, \(l = \pi r\)
Given that the radii are: \(r_1 = 0.5\text{ cm},\; r_2 = 1.0\text{ cm},\; r_3 = 1.5\text{ cm},\; \dots\)
Therefore, the lengths of the successive semicircles will be:
\(l_1 = 0.5\pi\)
\(l_2 = 1.0\pi\)
\(l_3 = 1.5\pi\)
\(\dots\)
These lengths form an Arithmetic Progression (AP):
\(0.5\pi,\; 1.0\pi,\; 1.5\pi,\; 2.0\pi,\; \dots\)
Here, first term, \(a = 0.5\pi\)
Common difference, \(d = 1.0\pi - 0.5\pi = 0.5\pi\)
Total number of semicircles, \(n = 13\)
The total length of the spiral is the sum of these 13 terms (\(S_{13}\)):
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_{13} = \frac{13}{2}[2(0.5\pi) + (13-1)0.5\pi]\)
\(\Rightarrow S_{13} = \frac{13}{2}[1.0\pi + 12 \times 0.5\pi]\)
\(\Rightarrow S_{13} = \frac{13}{2}[1\pi + 6\pi]\)
\(\Rightarrow S_{13} = \frac{13}{2} \times 7\pi\)
Now, substituting \(\pi = \frac{22}{7}\), we get:
\(\Rightarrow S_{13} = \frac{13}{2} \times 7 \times \frac{22}{7}\)
\(\Rightarrow S_{13} = \frac{13}{2} \times 22\)
\(\Rightarrow S_{13} = 13 \times 11\)
\(\Rightarrow S_{13} = 143\)
\(\therefore\) The total length of such a spiral is 143 cm.

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?

Solution:
The number of logs in the rows forms an Arithmetic Progression (AP):
\(20,\; 19,\; 18,\; 17,\; \dots\)
Here, first term, \(a = 20\)
Common difference, \(d = 19 - 20 = -1\)
Total number of logs (sum), \(S_n = 200\)
Let the total number of rows be \(= n\)
We know that, \(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow 200 = \frac{n}{2}[2(20) + (n-1)(-1)]\)
\(\Rightarrow 400 = n[40 - n + 1]\)
\(\Rightarrow 400 = n[41 - n]\)
\(\Rightarrow 400 = 41n - n^2\)
\(\Rightarrow n^2 - 41n + 400 = 0\)
By splitting the middle term, we get:
\(\Rightarrow n^2 - 16n - 25n + 400 = 0\)
\(\Rightarrow n(n - 16) - 25(n - 16) = 0\)
\(\Rightarrow (n - 16)(n - 25) = 0\)
\(\Rightarrow n = 16\) or \(n = 25\)

Verifying the conditions:
If \(n = 25\), then the number of logs in the top row will be:
\(a_{25} = a + 24d\)
\(\Rightarrow a_{25} = 20 + 24(-1) = 20 - 24 = -4\)
Since the number of logs cannot be negative, \(n = 25\) is not acceptable.

If \(n = 16\), then the number of logs in the top row will be:
\(a_{16} = a + 15d\)
\(\Rightarrow a_{16} = 20 + 15(-1) = 20 - 15 = 5\)
\(\therefore\) The \(200\) logs are placed in 16 rows and there are 5 logs in the top row.

20. In a potato race, a bucket is placed at the starting point, which is \(5\text{ m}\) from the first potato, and the other potatoes are placed \(3\text{ m}\) apart in a straight line. There are \(10\) potatoes in the line (see Fig. 5.6).

A competitor starts from the bucket, runs to pick 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 \(2\times5 + 2(5+3)\)]

Solution:
To pick up each potato and drop it into the bucket, the competitor has to cover twice the distance of the potato from the bucket.
Distance covered for the 1st potato \(= 2 \times 5 = 10\text{ m}\)
Distance covered for the 2nd potato \(= 2 \times (5 + 3) = 2 \times 8 = 16\text{ m}\)
Distance covered for the 3rd potato \(= 2 \times (5 + 3 + 3) = 2 \times 11 = 22\text{ m}\)
\(\dots\)
These distances form an Arithmetic Progression (AP):
\(10,\; 16,\; 22,\; \dots\)
Here, first term, \(a = 10\)
Common difference, \(d = 16 - 10 = 6\)
Total number of potatoes (number of terms), \(n = 10\)
The total distance the competitor has to run is the sum of these 10 terms (\(S_{10}\)):
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow S_{10} = \frac{10}{2}[2(10) + (10-1)6]\)
\(\Rightarrow S_{10} = 5[20 + 9 \times 6]\)
\(\Rightarrow S_{10} = 5[20 + 54]\)
\(\Rightarrow S_{10} = 5 \times 74\)
\(\Rightarrow S_{10} = 370\)
\(\bullet\) Therefore, the competitor has to run a total distance of 370 metres.

 

21. The first and the last terms of an Arithmetic Progression are \(1\) and \(11\) respectively. If the sum of its terms is \(36\), then the number of terms will be—

(A) \(5\)
(B) \(6\)
(C) \(7\)
(D) \(8\)

Solution:
Given that,
First term, \(a = 1\)
Last term, \(l = 11\)
Sum, \(S_n = 36\)
We know that,
\(S_n = \frac{n}{2}(a + l)\)
\(\Rightarrow 36 = \frac{n}{2}(1 + 11)\)
\(\Rightarrow 36 = \frac{n}{2}(12)\)
\(\Rightarrow 36 = 6n\)
\(\Rightarrow n = \frac{36}{6}\)
\(\Rightarrow n = 6\)
\(\therefore\) The correct option is (B).

22. Match the following columns.

List I	List II
P) Sum of first \(n\) natural numbers	1) \(n^2\)
Q) Sum up to \(n\) terms when first term is \(a\) and common difference is \(d\)	2) \(n(n+1)\)
R) Sum of first \(n\) odd natural numbers	3) \(\frac{n(n+1)}{2}\)
S) Sum of first \(n\) even natural numbers	4) \(\frac{n}{2}[2a+(n-1)d]\)

Choose the correct option—

(A) P → 2, Q → 3, R → 4, S → 1
(B) P → 4, Q → 2, R → 1, S → 3
(C) P → 3, Q → 4, R → 1, S → 2
(D) P → 2, Q → 1, R → 3, S → 4

Solution:
Verifying the mathematical formulas, we get:
P) Sum of first \(n\) natural numbers \(= \frac{n(n+1)}{2}\) \(\Rightarrow\) (3)
Q) Sum up to \(n\) terms of an AP \(= \frac{n}{2}[2a+(n-1)d]\) \(\Rightarrow\) (4)
R) Sum of first \(n\) odd natural numbers \(= n^2\) \(\Rightarrow\) (1)
S) Sum of first \(n\) even natural numbers \(= n(n+1)\) \(\Rightarrow\) (2)
Thus, the correct match is: P → 3, Q → 4, R → 1, S → 2.
\(\therefore\) The correct option is (C).

23. For the AP \(5,\;9,\;13,\;\ldots,\;185\), which of the following statements is correct?

(i) \(a_n=1+4n\)
(ii) \(a_3+a_4=30\)
(iii) \(S_n=\frac{n}{2}\times190\)
(iv) \(n=43\)

Choose the correct option—

(A) (i), (iii)    (B) (ii), (iii)    (C) (i), (ii), (iii)    (D) (i), (ii), (iii), (iv)

Solution:
Given AP: \(5, 9, 13, \dots, 185\)
Here, \(a = 5\) and \(d = 9 - 5 = 4\)

Verifying Statement (i):
\(a_n = a + (n-1)d\)
\(\Rightarrow a_n = 5 + (n-1)4\)
\(\Rightarrow a_n = 5 + 4n - 4\)
\(\Rightarrow a_n = 1 + 4n\) (Hence, statement i is correct)

Verifying Statement (iv):
Last term, \(a_n = 185\)
\(\Rightarrow 1 + 4n = 185\)
\(\Rightarrow 4n = 184\)
\(\Rightarrow n = 46\) (Hence, statement iv is incorrect since it is given as \(n=43\))

Verifying Statement (ii):
\(a_3 = 5 + 2(4) = 13\)
\(a_4 = 5 + 3(4) = 17\)
\(\Rightarrow a_3 + a_4 = 13 + 17 = 30\) (Hence, statement ii is correct)

Verifying Statement (iii):
\(S_n = \frac{n}{2}(a + a_n)\)
\(\Rightarrow S_n = \frac{n}{2}(5 + 185)\)
\(\Rightarrow S_n = \frac{n}{2} \times 190\) (Hence, statement iii is correct)

Since statements (i), (ii), and (iii) are correct, the right choice is (C).
\(\therefore\) The correct option is (C).

24. This question consists of an Assertion (A) and a Reason (R).

Assertion (A): The sum of the first \(10\) terms of the sequence \(11,\;13,\;15,\;17,\;\ldots\) is \(200\).
Reason (R): The sum of the first \(n\) odd natural numbers is \(n^2\).

Choose the correct option:

(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.

Solution:
Verifying Assertion (A):
Given sequence: \(11, 13, 15, 17, \dots\)
Here, \(a = 11, d = 2, n = 10\)
\(S_{10} = \frac{10}{2}[2(11) + (10-1)2]\)
\(\Rightarrow S_{10} = 5[22 + 9 \times 2]\)
\(\Rightarrow S_{10} = 5[22 + 18]\)
\(\Rightarrow S_{10} = 5 \times 40\)
\(\Rightarrow S_{10} = 200\) (Hence, Assertion A is true)

Verifying Reason (R):
The sum of the first \(n\) odd natural numbers is always \(n^2\) (this formula is universally true). However, the sequence given in Assertion (A) does not start from the first odd natural number \(1\); it starts from \(11\). Therefore, the formula given in Reason (R) cannot be applied directly to solve Assertion (A)—instead, we solved it using the general formula for the sum of an AP. Thus, Reason (R) is not the correct explanation of Assertion (A).
\(\therefore\) Both (A) and (R) are true but (R) is not the correct explanation of (A).
\(\therefore\) The correct option is (B).

25. If the sum of the first \(n\) odd numbers is \(S_1\) and the sum of the first \(n\) natural numbers is \(S_2\), then the value of \(\frac{S_1}{S_2}\) will be—

(A) \(\frac{2n}{n+1}\)     (B) \(\frac{n+1}{2n}\)     (C) \(\frac{n}{n+1}\)     (D) \(\frac{n+1}{n}\)

Solution:
We know that,
Sum of first \(n\) odd numbers, \(S_1 = n^2\)
Sum of first \(n\) natural numbers, \(S_2 = \frac{n(n+1)}{2}\)
Now, let's find the ratio:
\(\Rightarrow \frac{S_1}{S_2} = \frac{n^2}{\frac{n(n+1)}{2}}\)
\(\Rightarrow \frac{S_1}{S_2} = n^2 \times \frac{2}{n(n+1)}\)
\(\Rightarrow \frac{S_1}{S_2} = \frac{2n^2}{n(n+1)}\)
\(\Rightarrow \frac{S_1}{S_2} = \frac{2n}{n+1}\)
\(\therefore\) The correct option is (A).

26. A fruit seller arranges apples from a crate containing \(240\) apples such that there are \(6\) apples in the first row, \(10\) apples in the second row, \(14\) apples in the third row, and so on.

(i) Find the total number of rows of apples.
(ii) Find the difference between the number of apples in the seventh row and the third row.
(iii) Find the number of apples in the third row from the end.
(iv) Check whether there is any row containing \(32\) apples.

(i) Find the total number of rows of apples.

Solution:
The number of apples in the rows forms an AP: \(6, 10, 14, \dots\)
Here, first term \(a = 6\), common difference \(d = 10 - 6 = 4\)
Total number of apples (sum), \(S_n = 240\)
Let the total number of rows be \(= n\)
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\Rightarrow 240 = \frac{n}{2}[2(6) + (n-1)4]\)
\(\Rightarrow 240 \times 2 = n[12 + 4n - 4]\)
\(\Rightarrow 480 = n[4n + 8]\)
\(\Rightarrow 480 = 4n^2 + 8n\)
Dividing both sides by 4, we get:
\(\Rightarrow n^2 + 2n - 120 = 0\)
\(\Rightarrow n^2 + 12n - 10n - 120 = 0\)
\(\Rightarrow n(n + 12) - 10(n + 12) = 0\)
\(\Rightarrow (n - 10)(n + 12) = 0\)
\(\Rightarrow n = 10\) or \(n = -12\)
Since the number of rows cannot be negative, we take \(n = 10\).
\(\therefore\) The total number of rows of apples is 10.

(ii) Find the difference between the number of apples in the seventh row and the third row.

Solution:
Number of apples in the 7th row, \(a_7 = a + 6d\)
Number of apples in the 3rd row, \(a_3 = a + 2d\)
Required difference \(= a_7 - a_3\)
\(\Rightarrow a_7 - a_3 = (a + 6d) - (a + 2d)\)
\(\Rightarrow a_7 - a_3 = 4d\)
\(\Rightarrow a_7 - a_3 = 4 \times 4 = 16\)
\(\therefore\) The difference between the apples in the 7th row and the 3rd row is 16.

(iii) Find the number of apples in the third row from the end.

Solution:
Since the total number of rows is \(n = 10\), the third row from the end will be:
\(\Rightarrow 10 - 3 + 1 = 8\)-th row.
Now, let's find the number of apples in the 8th row (\(a_8\)):
\(a_8 = a + 7d\)
\(\Rightarrow a_8 = 6 + 7(4)\)
\(\Rightarrow a_8 = 6 + 28 = 34\)
\(\dots\) The number of apples in the third row from the end is 34.

(iv) Check whether there is any row containing \(32\) apples.

Solution:
Let us assume that the \(n\)-th row contains 32 apples.
\(\Rightarrow a_n = 32\)
\(\Rightarrow a + (n-1)d = 32\)
\(\Rightarrow 6 + (n-1)4 = 32\)
\(\Rightarrow 4(n-1) = 32 - 6\)
\(\Rightarrow 4(n-1) = 26\)
\(\Rightarrow n-1 = \frac{26}{4} = 6.5\)
\(\Rightarrow n = 6.5 + 1 = 7.5\)
Since the row number \(n\) must always be a positive integer, and here we obtained a fractional value, it is impossible to have a row containing exactly 32 apples.
\(\therefore\) There is no row containing 32 apples.

Sudev Chandra Das (B.Sc. Mathematics)

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Important Points for HSLC Exam

Students should remember:

Sum Formula
nth Term Formula
Common Difference
AP Word Problems
Formula-based Calculations

These topics are frequently asked in HSLC board examinations.

Real-Life Uses of Arithmetic Progression

Arithmetic Progression is used in:

Banking and finance
Salary increment calculation
Monthly savings
Business mathematics
Computer programming
Engineering calculations

Benefits of These Solutions

✅ Based on latest SEBA/SCERT Assam 2026–2027 syllabus
✅ Easy English explanations
✅ Step-by-step solutions
✅ HSLC exam focused preparation
✅ Important formulas included
✅ Student-friendly learning method

Final Words

These Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 Solutions help students understand AP formulas and calculations in a simple and easy way. Regular practice of all questions will improve mathematical skills and help students score high marks in HSLC Examination.

Students should practice:

Sum Formula
nth Term Formula
AP Pattern Questions
Formula-based Problems
Word Problems

daily for better understanding and strong mathematical foundation.

Class 10 Maths Chapter 5 – Arithmetic Progressions (Exercise 5.3) FAQs

Exercise 5.3 focuses on finding the sum of the first n terms of an Arithmetic Progression (AP).

👉 Students learn how to calculate the sum quickly using the AP sum formula.

The formula for the sum of the first n terms of an Arithmetic Progression is:

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📘 Here, a = first term, d = common difference, and n = number of terms.

Yes, Exercise 5.3 is very important for HSLC exams. Questions worth 3–5 marks are frequently asked from this topic.

🎯 Proper use of the formula can help students score full marks easily.

The difference between two consecutive terms of an Arithmetic Progression is called the common difference (d).

Questions include finding the sum of AP terms, identifying APs, and solving real-life application problems.

Students often make mistakes while substituting values of a, d, or n into the formula.

🚀 Pro Tip: Carefully identify all values before applying the formula.

Regular practice, formula revision, and solving previous year question papers are very important for scoring high marks.

⭐ Daily Maths practice improves speed, accuracy, and confidence.

Digital Pipal Academy provides step-by-step solutions, updated syllabus content, easy explanations, and exam-oriented preparation.

🎓 It helps students improve concepts, accuracy, and confidence in Mathematics.

You can find complete step-by-step solutions for Exercise 5.3 on Digital Pipal Academy.