SEBA Class 10 Maths Exercise 5.1 Solutions: Arithmetic Progressions | New Book 2026
Get free solutions to SEBA Class 10 Maths Exercise 5.1 Solutions from the new SCERT textbook 2026. In this article, we have provided basic concepts of an Arithmetic Progression (AP), the common difference, the general form of an AP, finite AP, and infinite AP. Also, we solved all questions, including MCQs, in a simple, step-by-step manner.
If you are a Class 10 student under SEBA/Assam Board preparing for your upcoming HSLC examinations, these reliable, easy-to-understand solutions will help you clear your concepts and score full marks in your exams.
Chapter 5: Arithmetic Progressions | Basic Notes
1. Arithmetic Progression (AP)
An Arithmetic Progression (AP) is a sequence or list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term.
- Example: $1, 2, 3, 4, \dots$
Each number in the sequence is called a term.
2. Common Difference ($d$)
The fixed number added to each term is called the common difference of the AP. It is denoted by $d$.
The common difference can be positive, negative, or zero.
For an AP with terms $a_1, a_2, a_3, \dots, a_n$:
$$d = a_2 – a_1 = a_3 – a_2 = \dots = a_{k+1} – a_k$$
3. General Form of an AP
If $a$ is the first term and $d$ is the common difference, the general form of an AP is written as:
First term ($a_1$): $a$
Second term ($a_2$): $a + d$
Third term ($a_3$): $a + 2d$
Fourth term ($a_4$): $a + 3d$
4. Finite vs. Infinite AP
Finite AP: An AP containing a finite number of terms is called a finite AP. Every finite AP has a last term.
Example: $147, 148, 149, \dots, 157$
Infinite AP: An AP containing an endless number of terms is called an infinite AP. These APs do not have a last term.
Example: $1, 2, 3, 4, \dots$
See More:
SEBA Class 10 Maths Exercise 5.1 Solutions: Arithmetic Progressions
SEBA Class 10 Maths Exercise 5.2 Solutions: Arithmetic Progressions
SEBA Class 10 Maths Exercise 5.3 Solutions: Arithmetic Progressions
EXERCISE 5.1- Arithmetic Progressions
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 Rs 15 for the first km and Rs 8 for each additional km.
$\mathbf{Sol^n.}$
Given:
Fare for the $1^{\text{st}}$ km ($a_1$) = Rs 15
Fare for the $2^{\text{nd}}$ km ($a_2$) = Rs $15 + 8 = \text{Rs } 23$
Fare for the $3^{\text{rd}}$ km ($a_3$) = Rs $23 + 8 = \text{Rs } 31$
Fare for the $4^{\text{th}}$ km ($a_4$) = Rs $31 + 8 = \text{Rs } 39$
Now,
$a_2 – a_1 = 23 – 15 = 8$
$a_3 – a_2 = 31 – 23 = 8$
$a_4 – a_3 = 39 – 31 = 8$
Since $a_{k+1} – a_k$ is the same every time ($d = 8$), the given situation forms an Arithmetic Progression (AP).
(ii) The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.
$\mathbf{Sol^n.}$
Let the initial volume of air in the cylinder be $V$.
Air present initially ($a_1$) = $V$
Air removed in the $1^{\text{st}}$ stroke = $\frac{1}{4}V$
Air remaining after $1^{\text{st}}$ stroke ($a_2$) = $V – \frac{1}{4}V = \frac{3}{4}V$
Air removed in the $2^{\text{nd}}$ stroke = $\frac{1}{4}\left(\frac{3}{4}V\right) = \frac{3}{16}V$
Air remaining after $2^{\text{nd}}$ stroke ($a_3$) = $\frac{3}{4}V – \frac{3}{16}V = \frac{9}{16}V = \left(\frac{3}{4}\right)^2 V$
Now,
$a_2 – a_1 = \frac{3}{4}V – V = -\frac{1}{4}V$
$a_3 – a_2 = \frac{9}{16}V – \frac{3}{4}V = \frac{9V – 12V}{16} = -\frac{3}{16}V$
Since $a_2 – a_1 \ne a_3 – a_2$, the given situation does not form an AP.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
$\mathbf{Sol^n.}$
Cost of digging for the $1^{\text{st}}$ metre ($a_1$) = Rs 150
Cost of digging for 2 metres ($a_2$) = Rs $150 + 50 = \text{Rs } 200$
Cost of digging for 3 metres ($a_3$) = Rs $200 + 50 = \text{Rs } 250$
Cost of digging for 4 metres ($a_4$) = Rs $250 + 50 = \text{Rs } 300$
Now,
$a_2 – a_1 = 200 – 150 = 50$
$a_3 – a_2 = 250 – 200 = 50$
$a_4 – a_3 = 300 – 250 = 50$
Thus, $a_4 – a_3 = a_3 – a_2 = a_2 – a_1$
So, the given situation forms an Arithmetic Progression (AP).
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
$\mathbf{Sol^n.}$
Given the initial deposit, $a_1 = 10000$
Amount of money after 1 year, $a_2 = 10000 + 10000 \times \frac{8}{100}$
$= 10000 + 800$
$= 10800$
Amount of money after 2 years, $a_3 = 10800 + 10800 \times \frac{8}{100}$
$= 10800 + 864$
$= 11664$
Amount of money after 3 years, $a_4 = 11664 + 11664 \times \frac{8}{100}$
$= 11664 + 933.12$
$= 12597.12$
Now,
$a_2 – a_1 = 10800 – 10000 = 800$
$a_3 – a_2 = 11664 – 10800 = 864$
Since $a_2 – a_1 \ne a_3 – a_2$, the given situation does not form an AP.
2. Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:
(i) $a = 10, \; d = 10$
$\mathbf{Sol^n.}$
$a_1 = a = 10$
$a_2 = a + d = 10 + 10 = 20$
$a_3 = a_2 + d = 20 + 10 = 30$
$a_4 = a_3 + d = 30 + 10 = 40$
Hence, the first four terms are 10, 20, 30, 40.
(ii) $a = -2, \; d = 0$
$\mathbf{Sol^n.}$
$a_1 = a = -2$
$a_2 = a + d = -2 + 0 = -2$
$a_3 = a_2 + d = -2 + 0 = -2$
$a_4 = a_3 + d = -2 + 0 = -2$
Hence, the first four terms of the AP are -2, -2, -2, -2.
(iii) $a = 4, \; d = -3$
$\mathbf{Sol^n.}$
$a_1 = a = 4$
$a_2 = a + d = 4 + (-3) = 1$
$a_3 = a_2 + d = 1 + (-3) = -2$
$a_4 = a_3 + d = -2 + (-3) = -5$
Hence, the first four terms are 4, 1, -2, -5.
(iv) $a = -1, \; d = \frac{1}{2}$
$\mathbf{Sol^n.}$
$a_1 = a = -1$
$a_2 = a + d = -1 + \frac{1}{2} = -\frac{1}{2}$
$a_3 = a_2 + d = -\frac{1}{2} + \frac{1}{2} = 0$
$a_4 = a_3 + d = 0 + \frac{1}{2} = \frac{1}{2}$
Hence, the first four terms are $-1, -\frac{1}{2}, 0, \frac{1}{2}$.
(v) $a = -1.25, \; d = -0.25$
$\mathbf{Sol^n.}$
$a_1 = a = -1.25$
$a_2 = a + d = -1.25 + (-0.25) = -1.50$
$a_3 = a_2 + d = -1.50 + (-0.25) = -1.75$
$a_4 = a_3 + d = -1.75 + (-0.25) = -2.00$
Hence, the first four terms are -1.25, -1.50, -1.75, -2.00.
3. For the following APs, write the first term and the common difference:
(i) $3, 1, -1, -3, \dots$
$\mathbf{Sol^n.}$
First term ($a$) = $3$
Common difference ($d$) = $a_2 – a_1 = 1 – 3 = -2$
(ii) $-5, -1, 3, 7, \dots$
$\mathbf{Sol^n.}$
First term ($a$) = $-5$
Common difference ($d$) = $a_2 – a_1 = -1 – (-5) = -1 + 5 = 4$
(iii) $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$
$\mathbf{Sol^n.}$
First term ($a$) = $\frac{1}{3}$
Common difference ($d$) = $a_2 – a_1 = \frac{5}{3} – \frac{1}{3} = \frac{4}{3}$
(iv) $0.6, 1.7, 2.8, 3.9, \dots$
$\mathbf{Sol^n.}$
First term ($a$) = $0.6$
Common difference ($d$) = $a_2 – a_1 = 1.7 – 0.6 = 1.1$
4. Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms:
(i) $2, 4, 8, 16, \dots$
$\mathbf{Sol^n.}$
$a_2 – a_1 = 4 – 2 = 2$
$a_3 – a_2 = 8 – 4 = 4$
Since $a_2 – a_1 \ne a_3 – a_2$, the given list of numbers does not form an AP.
(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
$\mathbf{Sol^n.}$
We have,
$a_2 – a_1 = \frac{5}{2} – 2 = \frac{1}{2}$
$a_3 – a_2 = 3 – \frac{5}{2} = \frac{1}{2}$
$a_4 – a_3 = \frac{7}{2} – 3 = \frac{1}{2}$
Thus, $a_{k+1} – a_k$ is the same every time.
So, the given list of numbers forms an AP with the common difference $d = \frac{1}{2}$
The next three terms are:
$a_5 = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$
$a_6 = 4 + \frac{1}{2} = \frac{9}{2}$
$a_7 = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5$
(iii) $-1.2, -3.2, -5.2, -7.2, \dots$
$\mathbf{Sol^n.}$
We have,
$a_2 – a_1 = -3.2 – (-1.2) = -2$
$a_3 – a_2 = -5.2 – (-3.2) = -2$
$a_4 – a_3 = -7.2 – (-5.2) = -2$
Thus, $a_{k+1} – a_k$ is the same every time.
So, the given list of numbers forms an AP with the common difference $d = -2$.
The next three terms are:
$a_5 = -7.2 + (-2) = -9.2$
$a_6 = -9.2 + (-2) = -11.2$
$a_7 = -11.2 + (-2) = -13.2$
(iv) $-10, -6, -2, 2, \dots$
$\mathbf{Sol^n.}$
We have,
$a_2 – a_1 = -6 – (-10) = 4$
$a_3 – a_2 = -2 – (-6) = 4$
$a_4 – a_3 = 2 – (-2) = 4$
So, the given list of numbers forms an AP with the common difference $d = 4$.
The next three terms are:
$a_5 = 2 + 4 = 6$
$a_6 = 6 + 4 = 10$
$a_7 = 10 + 4 = 14$
(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$
$\mathbf{Sol^n.}$
We have,
$a_2 – a_1 = (3+\sqrt{2}) – 3 = \sqrt{2}$
$a_3 – a_2 = (3+2\sqrt{2}) – (3+\sqrt{2}) = \sqrt{2}$
$a_4 – a_3 = (3+3\sqrt{2}) – (3+2\sqrt{2}) = \sqrt{2}$
So, the given list of numbers forms an AP with the common difference $d = \sqrt{2}$.
The next three terms are:
$a_5 = (3+3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$
$a_6 = (3+4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$
$a_7 = (3+5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$
(vi) $0.2, 0.22, 0.222, 0.2222, \dots$
$\mathbf{Sol^n.}$
We have,
$a_2 – a_1 = 0.22 – 0.2 = 0.02$
$a_3 – a_2 = 0.222 – 0.22 = 0.002$
As $a_2 – a_1 \ne a_3 – a_2$, the given list of numbers does not form an AP.
(vii) $0, -4, -8, -12, \dots$
$\mathbf{Sol^n.}$
We have,
$a_2 – a_1 = -4 – 0 = -4$
$a_3 – a_2 = -8 – (-4) = -4$
$a_4 – a_3 = -12 – (-8) = -4$
So, the given list of numbers forms an AP with the common difference $d = -4$.
The next three terms are:
$a_5 = -12 + (-4) = -16$
$a_6 = -16 + (-4) = -20$
$a_7 = -20 + (-4) = -24$
(viii) $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$
$\mathbf{Sol^n.}$
We have
$a_2 – a_1 = -\frac{1}{2} – \left(-\frac{1}{2}\right) = 0$
$a_3 – a_2 = -\frac{1}{2} – \left(-\frac{1}{2}\right) = 0$
$a_4 – a_3 = -\frac{1}{2} – \left(-\frac{1}{2}\right) = 0$
Thus, the given sequence is an AP with common difference $d = 0$.
Next three terms are:
$a_5 = -\frac{1}{2}$
$a_6 = -\frac{1}{2}$
$a_7 = -\frac{1}{2}$
(ix) $1, 3, 9, 27, \dots$
$\mathbf{Sol^n.}$
$a_2 – a_1 = 3 – 1 = 2$
$a_3 – a_2 = 9 – 3 = 6$
As $a_2 – a_1 \ne a_3 – a_2$, the given list of numbers does not form an AP.
(x) $a, 2a, 3a, 4a, \dots$
$\mathbf{Sol^n.}$
$a_2 – a_1 = 2a – a = a$
$a_3 – a_2 = 3a – 2a = a$
$a_4 – a_3 = 4a – 3a = a$
Thus, the given sequence is an AP with common difference $d = a$.
Next three terms are:
$a_5 = 4a + a = 5a$
$a_6 = 5a + a = 6a$
$a_7 = 6a + a = 7a$
(xi) $a, a^2, a^3, a^4, \dots$
$\mathbf{Sol^n.}$
$a_2 – a_1 = a^2 – a = a(a – 1)$
$a_3 – a_2 = a^3 – a^2 = a^2(a – 1)$
As $a_2 – a_1 \ne a_3 – a_2$, the given list of numbers does not form an AP.
(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$
$\mathbf{Sol^n.}$
Rewriting the terms, we have:
$\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$
$\sqrt{2}, \; 2\sqrt{2}, \; 3\sqrt{2}, \; 4\sqrt{2}, \dots$
$a_2 – a_1 = 2\sqrt{2} – \sqrt{2} = \sqrt{2}$
$a_3 – a_2 = 3\sqrt{2} – 2\sqrt{2} = \sqrt{2}$
$a_4 – a_3 = 4\sqrt{2} – 3\sqrt{2} = \sqrt{2}$
So, the given list of numbers forms an AP with the common difference $d = \sqrt{2}$.
Next three terms are:
$a_5 = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25\times2} = \sqrt{50}$
$a_6 = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72}$
$a_7 = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98}$
(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$
$\mathbf{Sol^n.}$
$a_2 – a_1 = \sqrt{6} – \sqrt{3} = \sqrt{3}(\sqrt{2} – 1)$
$a_3 – a_2 = \sqrt{9} – \sqrt{6} = 3 – \sqrt{6}$
As $a_2 – a_1 \ne a_3 – a_2$, the given list of numbers does not form an AP.
(xiv) $1^2, 3^2, 5^2, 7^2, \dots$
$\mathbf{Sol^n.}$
Rewriting terms: $1, 9, 25, 49, \dots$
$a_2 – a_1 = 9 – 1 = 8$
$a_3 – a_2 = 25 – 9 = 16$
As $a_2 – a_1 \ne a_3 – a_2$, the given list of numbers does not form an AP.
(xv) $1^2, 5^2, 7^2, 73, \dots$
$\mathbf{Sol^n.}$
Rewriting the terms, we have:
$1, 25, 49, 73, \dots$
$a_2 – a_1 = 25 – 1 = 24$
$a_3 – a_2 = 49 – 25 = 24$
$a_4 – a_3 = 73 – 49 = 24$
So, the given list of numbers forms an AP with the common difference $d = 24$.
Next three terms are:
$a_5 = 73 + 24 = 97$
$a_6 = 97 + 24 = 121$
$a_7 = 121 + 24 = 145$
5. Which of the following list of numbers does not form an A.P.?
(i) $2, 4, 6, 8, 10, \dots$
(ii) $2, 4, 8, 16, \dots$
(iii) $1, -2, -5, -8, \dots$
(iv) $1.8, 2.0, 2.2, 2.4, \dots$
(A) (i) (B) (ii) (C) (iii) (D) (iv)
$\mathbf{Sol^n.}$
In (i): $4 – 2 = 6 – 4 = 2$ (Forms an AP)
In (ii): $4 – 2 = 2$, but $8 – 4 = 4 \ne 2$ (Does not form an AP)
In (iii): $-2 – 1 = -3$, $-5 – (-2) = -3$ (Forms an AP)
In (iv): $2.0 – 1.8 = 0.2$, $2.2 – 2.0 = 0.2$ (Forms an AP)
So, the correct option is (B)
6. Consider the statements P to S. They form jumbled links of common difference in A.P. Choose the option that represents the correct increasing sequence of common difference of the A.P.
$\text{P. } 2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
$\text{Q. } 1.2, 3.2, 5.2, 7.2, \dots$
$\text{R. } 7, 10\frac{1}{2}, 14, 17\frac{1}{2}, \dots$
$\text{S. } \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \frac{7}{60}, \dots$
(A) S, Q, R, P
(B) R, P, Q, S
(C) S, P, Q, R
(D) P, Q, S, R
$\mathbf{Sol^n.}$
Here, we calculate the common difference ($d$) for each:
For P: $d = \frac{5}{2} – 2 = \frac{1}{2} = 0.5$
For Q: $d = 3.2 – 1.2 = 2$
For R: $d = 10\frac{1}{2} – 7 = \frac{21}{2} – 7 = \frac{7}{2} = 3.5$
For S: $d = \frac{1}{12} – \frac{1}{15} = \frac{5 – 4}{60} = \frac{1}{60} \approx 0.0167$
Now, arranging the common differences in increasing order, we get:
$0.0167 < 0.5 < 2 < 3.5$
$\therefore \text{S} < \text{P} < \text{Q} < \text{R}$
So, the correct option is (C)
7. If $p, 2p – 1$ and $2p + 1$ are three consecutive terms of an A.P., the value of $p$ is:
(A) -3 (B) -2 (C) 3 (D) 6
$\mathbf{Sol^n.}$
If $a, b, c$ are in AP, then $b – a = c – b $
$\implies 2b = a + c$.
Here, $a = p$, $b = 2p – 1$, and $c = 2p + 1$.
Now, $2(2p – 1) = p + (2p + 1)$
$\implies 4p – 2 = 3p + 1$
$\implies 4p – 3p = 1 + 2$
$\implies p = 3$
So, the correct option is (C)
8. Show that $a – b, \; a, \; a + b$ form consecutive terms of an A.P.
$\mathbf{Sol^n.}$
Let the given terms be $t_1 = a – b$, $t_2 = a$, and $t_3 = a + b$.
Now,
$t_2 – t_1 = a – (a – b) = a – a + b = b$
$t_3 – t_2 = (a + b) – a = b$
Since $t_2 – t_1 = t_3 – t_2 = b$, the given three terms $a – b, \; a, \; a + b$ form consecutive terms of an Arithmetic Progression (AP).