CBSE Class 10 · Maths
CBSE Class 10 Maths Arithmetic Progressions Previous Year Questions
Help your child master CBSE Class 10 Maths Arithmetic Progressions Previous Year Questions with this curated collection sourced from real board papers spanning 2018–2025. Every question comes with a detailed step-by-step solution, helping your child confidently tackle the nth term formula, sum of terms, and real-life AP problems — topics that consistently carry marks across all question types in the board exam.
CBSE Class 10 Maths Arithmetic Progressions — Questions with Solutions
The next term of the A.P. $\sqrt{6}, \sqrt{24}, \sqrt{54}, \ldots$ is:
Solution
Answer: Option (B) is correct.
Explanation:
First term, $a_1 = \sqrt{6}$
Second term, $a_2 = \sqrt{24} = 2\sqrt{6}$
Third term, $a_3 = \sqrt{54} = 3\sqrt{6}$
Common difference $= \sqrt{6}(2 – 1) = \sqrt{6}$
Next (4th) term $= 3\sqrt{6} + \sqrt{6} = 4\sqrt{6} = \sqrt{96}$
Explanation:
First term, $a_1 = \sqrt{6}$
Second term, $a_2 = \sqrt{24} = 2\sqrt{6}$
Third term, $a_3 = \sqrt{54} = 3\sqrt{6}$
Common difference $= \sqrt{6}(2 – 1) = \sqrt{6}$
Next (4th) term $= 3\sqrt{6} + \sqrt{6} = 4\sqrt{6} = \sqrt{96}$
If $k + 2$, $4k – 6$, and $3k – 2$ are three consecutive terms of an A.P., then the value of $k$ is:
Solution
Answer: Option (A) is correct.
Explanation: Since $k + 2$, $4k – 6$, and $3k – 2$ are consecutive terms of an A.P.:
$(4k – 6) – (k + 2) = (3k – 2) – (4k – 6)$
$3k – 8 = -k + 4$
$4k = 12$
$k = 3$
Explanation: Since $k + 2$, $4k – 6$, and $3k – 2$ are consecutive terms of an A.P.:
$(4k – 6) – (k + 2) = (3k – 2) – (4k – 6)$
$3k – 8 = -k + 4$
$4k = 12$
$k = 3$
If the sum of the first $n$ terms of an A.P. is $3n^2 + n$ and its common difference is 6, then its first term is:
Solution
Answer: Option (D) is correct.
Explanation:
$S_n = 3n^2 + n$ and $d = 6$
Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$:
$3n^2 + n = \dfrac{n}{2}[2a + 6n – 6]$
$6n^2 + 2n = 2an + 6n^2 – 6n$
$8n = 2an \Rightarrow a = 4$
Therefore, the first term $= 4$.
Explanation:
$S_n = 3n^2 + n$ and $d = 6$
Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$:
$3n^2 + n = \dfrac{n}{2}[2a + 6n – 6]$
$6n^2 + 2n = 2an + 6n^2 – 6n$
$8n = 2an \Rightarrow a = 4$
Therefore, the first term $= 4$.
Two APs have the same common difference. The first term of one of these is $-1$ and that of the other is $-8$. The difference between their 4th terms is:
Solution
Answer: Option (C) is correct.
Explanation: Let both APs have common difference $d$.
4th term of first A.P.: $T_4 = -1 + 3d$
4th term of second A.P.: $T_4′ = -8 + 3d$
$T_4 – T_4′ = (-1 + 3d) – (-8 + 3d) = 7$
Hence, the difference between their 4th terms is $\mathbf{7}$.
Explanation: Let both APs have common difference $d$.
4th term of first A.P.: $T_4 = -1 + 3d$
4th term of second A.P.: $T_4′ = -8 + 3d$
$T_4 – T_4′ = (-1 + 3d) – (-8 + 3d) = 7$
Hence, the difference between their 4th terms is $\mathbf{7}$.
If the sum of first $n$ odd natural numbers is equal to $k$ times the sum of first $n$ even natural numbers, then $k$ is equal to:
Solution
Answer: Option (D) is correct.
Explanation:
Sum of first $n$ odd natural numbers $= n^2$
Sum of first $n$ even natural numbers $= n(n + 1)$
$n^2 = k \times n(n + 1)$
$\Rightarrow k = \dfrac{n}{n + 1}$
Explanation:
Sum of first $n$ odd natural numbers $= n^2$
Sum of first $n$ even natural numbers $= n(n + 1)$
$n^2 = k \times n(n + 1)$
$\Rightarrow k = \dfrac{n}{n + 1}$
The first term of an A.P. is $p$ and the common difference is $q$. Then its 10th term is:
Solution
Answer: Option (C) is correct.
Explanation: $a = p$ and $d = q$
10th term $= a + (10 – 1)d = p + 9q$
Explanation: $a = p$ and $d = q$
10th term $= a + (10 – 1)d = p + 9q$
Which of the following is not an A.P.?
Solution
Answer: Option (C) is correct.
Explanation: Checking Option (C): $\dfrac{4}{3},\ \dfrac{7}{3},\ \dfrac{9}{3},\ \dfrac{12}{3},\ \ldots$
$\dfrac{7}{3} – \dfrac{4}{3} = 1$
$\dfrac{9}{3} – \dfrac{7}{3} = \dfrac{2}{3}$
$\dfrac{12}{3} – \dfrac{9}{3} = 1$
The common differences are not equal. Hence, it is not an A.P.
Explanation: Checking Option (C): $\dfrac{4}{3},\ \dfrac{7}{3},\ \dfrac{9}{3},\ \dfrac{12}{3},\ \ldots$
$\dfrac{7}{3} – \dfrac{4}{3} = 1$
$\dfrac{9}{3} – \dfrac{7}{3} = \dfrac{2}{3}$
$\dfrac{12}{3} – \dfrac{9}{3} = 1$
The common differences are not equal. Hence, it is not an A.P.
4 groups in a class were asked to come up with an arithmetic progression (AP). Shown below are their responses:
Which of these groups correctly came up with an AP?
| Group | Arithmetic Progression |
|---|---|
| M | $4,\ 2,\ 0,\ -2,\ \ldots$ |
| N | $41,\ 38.5,\ 36,\ 33.5,\ \ldots$ |
| O | $-19,\ -21,\ -23,\ -25,\ \ldots$ |
| P | $-3,\ -3,\ -3,\ -3,\ \ldots$ |
Solution
Answer: Option (D) is correct.
Explanation:
Group M: $d = 2 – 4 = 0 – 2 = -2 – 0 = -2$ (constant) ✓
Group N: $d = 38.5 – 41 = 36 – 38.5 = 33.5 – 36 = -2.5$ (constant) ✓
Group O: $d = -21 – (-19) = -23 – (-21) = -2$ (constant) ✓
Group P: $d = -3 – (-3) = 0$ (constant) ✓
All groups M, N, O, and P have a constant common difference, so all are valid A.P.s.
Explanation:
Group M: $d = 2 – 4 = 0 – 2 = -2 – 0 = -2$ (constant) ✓
Group N: $d = 38.5 – 41 = 36 – 38.5 = 33.5 – 36 = -2.5$ (constant) ✓
Group O: $d = -21 – (-19) = -23 – (-21) = -2$ (constant) ✓
Group P: $d = -3 – (-3) = 0$ (constant) ✓
All groups M, N, O, and P have a constant common difference, so all are valid A.P.s.
The common difference of the A.P. whose $n^{\text{th}}$ term is given by $a_n = 3n + 7$, is:
Solution
Answer: Option (B) is correct.
Explanation:
$a_1 = 3(1) + 7 = 10$
$a_2 = 3(2) + 7 = 13$
$d = a_2 – a_1 = 13 – 10 = 3$
In the general form $a_n = an + b$, the common difference equals the coefficient of $n$, i.e., $d = 3$.
Explanation:
$a_1 = 3(1) + 7 = 10$
$a_2 = 3(2) + 7 = 13$
$d = a_2 – a_1 = 13 – 10 = 3$
In the general form $a_n = an + b$, the common difference equals the coefficient of $n$, i.e., $d = 3$.
If $p – 1$, $p + 1$, and $2p + 3$ are in A.P., then the value of $p$ is:
Solution
Answer: Option (C) is correct.
Explanation: For A.P., $a_2 – a_1 = a_3 – a_2$:
$(p + 1) – (p – 1) = (2p + 3) – (p + 1)$
$2 = p + 2$
$p = 0$
Explanation: For A.P., $a_2 – a_1 = a_3 – a_2$:
$(p + 1) – (p – 1) = (2p + 3) – (p + 1)$
$2 = p + 2$
$p = 0$
The value of $x$ for which $2x$, $(x + 10)$, and $(3x + 2)$ are the three consecutive terms of an A.P. is:
Solution
Answer: Option (A) is correct.
Explanation: For A.P., the difference between consecutive terms must be equal:
$(x + 10) – 2x = (3x + 2) – (x + 10)$
$10 – x = 2x – 8$
$18 = 3x$
$x = 6$
Explanation: For A.P., the difference between consecutive terms must be equal:
$(x + 10) – 2x = (3x + 2) – (x + 10)$
$10 – x = 2x – 8$
$18 = 3x$
$x = 6$
If the sum of the first $m$ terms of an A.P. is given by $S_m = 2m^2 + 3m$, then the second term of the A.P. is:
Solution
Answer: Option (B) is correct.
Explanation: Using $a_n = S_n – S_{n-1}$:
$S_2 = 2(4) + 3(2) = 8 + 6 = 14$
$S_1 = 2(1) + 3(1) = 5$
$a_2 = S_2 – S_1 = 14 – 5 = \mathbf{9}$
Explanation: Using $a_n = S_n – S_{n-1}$:
$S_2 = 2(4) + 3(2) = 8 + 6 = 14$
$S_1 = 2(1) + 3(1) = 5$
$a_2 = S_2 – S_1 = 14 – 5 = \mathbf{9}$
In an A.P., if $a = 8$ and $a_{10} = -19$, then the value of $d$ is:
Solution
Answer: Option (D) is correct.
Explanation: Using $a_n = a + (n-1)d$:
$a_{10} = 8 + 9d = -19$
$9d = -27$
$d = -3$
Explanation: Using $a_n = a + (n-1)d$:
$a_{10} = 8 + 9d = -19$
$9d = -27$
$d = -3$
In an A.P., if the first term $a = 7$, the $n$-th term $a_n = 84$, and the sum of the first $n$ terms $S_n = \dfrac{2093}{2}$, then $n$ is equal to:
Solution
Answer: Option (C) is correct.
Explanation: Using $S_n = \dfrac{n}{2}(a + a_n)$:
$\dfrac{2093}{2} = \dfrac{n}{2}(7 + 84) = \dfrac{91n}{2}$
$2093 = 91n \Rightarrow n = \dfrac{2093}{91} = 23$
Explanation: Using $S_n = \dfrac{n}{2}(a + a_n)$:
$\dfrac{2093}{2} = \dfrac{n}{2}(7 + 84) = \dfrac{91n}{2}$
$2093 = 91n \Rightarrow n = \dfrac{2093}{91} = 23$
If $a$, $b$, $c$ form an A.P. with common difference $d$, then the value of $a + 2b – c$ is equal to:
Solution
Answer: Option (C) is correct.
Explanation: Since $a$, $b$, $c$ are in A.P. with common difference $d$:
$b = a + d$ and $c = a + 2d$
$a + 2b – c = a + 2(a + d) – (a + 2d)$
$= a + 2a + 2d – a – 2d$
$= 2a$
(Note: The value simplifies to $2a$; verify your answer choice against your question paper.)
Explanation: Since $a$, $b$, $c$ are in A.P. with common difference $d$:
$b = a + d$ and $c = a + 2d$
$a + 2b – c = a + 2(a + d) – (a + 2d)$
$= a + 2a + 2d – a – 2d$
$= 2a$
(Note: The value simplifies to $2a$; verify your answer choice against your question paper.)
Show that $(a – b)^2$, $(a^2 + b^2)$, and $(a + b)^2$ are in A.P.
Answer
Given: $(a – b)^2$, $(a^2 + b^2)$, $(a + b)^2$
Common difference $d_1 = (a^2 + b^2) – (a – b)^2$
$= a^2 + b^2 – (a^2 – 2ab + b^2)$
$= a^2 + b^2 – a^2 – b^2 + 2ab = 2ab$
Common difference $d_2 = (a + b)^2 – (a^2 + b^2)$
$= a^2 + 2ab + b^2 – a^2 – b^2 = 2ab$
Since $d_1 = d_2 = 2ab$, the three terms form an A.P. Hence Proved.
Common difference $d_1 = (a^2 + b^2) – (a – b)^2$
$= a^2 + b^2 – (a^2 – 2ab + b^2)$
$= a^2 + b^2 – a^2 – b^2 + 2ab = 2ab$
Common difference $d_2 = (a + b)^2 – (a^2 + b^2)$
$= a^2 + 2ab + b^2 – a^2 – b^2 = 2ab$
Since $d_1 = d_2 = 2ab$, the three terms form an A.P. Hence Proved.
Find $a$ and $b$ so that the numbers $a$, 7, $b$, 23 are in A.P.
Answer
Given: $a$, 7, $b$, 23 are in A.P.
Since terms of an A.P. have equal common differences:
$7 – a = b – 7 = 23 – b$
Step 1: From $b – 7 = 23 – b$:
$2b = 30 \Rightarrow b = 15$
Step 2: From $7 – a = b – 7 = 15 – 7 = 8$:
$7 – a = 8 \Rightarrow a = -1$
Therefore, $a = -1$ and $b = 15$.
Since terms of an A.P. have equal common differences:
$7 – a = b – 7 = 23 – b$
Step 1: From $b – 7 = 23 – b$:
$2b = 30 \Rightarrow b = 15$
Step 2: From $7 – a = b – 7 = 15 – 7 = 8$:
$7 – a = 8 \Rightarrow a = -1$
Therefore, $a = -1$ and $b = 15$.
How many terms are there in an A.P. whose first and fifth terms are $-14$ and $2$, respectively, and the last term is $62$?
Answer
Given: $a = -14$, $a_5 = 2$, $a_n = 62$
Step 1: Find the common difference $d$:
$a_5 = a + 4d \Rightarrow 2 = -14 + 4d \Rightarrow d = 4$
Step 2: Find $n$ using $a_n = a + (n-1)d$:
$62 = -14 + (n – 1) \times 4$
$76 = 4(n – 1)$
$n – 1 = 19 \Rightarrow n = 20$
There are 20 terms in the A.P.
Step 1: Find the common difference $d$:
$a_5 = a + 4d \Rightarrow 2 = -14 + 4d \Rightarrow d = 4$
Step 2: Find $n$ using $a_n = a + (n-1)d$:
$62 = -14 + (n – 1) \times 4$
$76 = 4(n – 1)$
$n – 1 = 19 \Rightarrow n = 20$
There are 20 terms in the A.P.
Find the sum of first 20 terms of the following A.P.: $1, 4, 7, 10, \ldots$
Answer
Given A.P.: $1, 4, 7, 10, \ldots$
$a = 1$, $d = 3$, $n = 20$
Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$:
$S_{20} = \dfrac{20}{2}[2 \times 1 + (20-1) \times 3]$
$= 10(2 + 57)$
$= 10 \times 59 = \boxed{590}$
$a = 1$, $d = 3$, $n = 20$
Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$:
$S_{20} = \dfrac{20}{2}[2 \times 1 + (20-1) \times 3]$
$= 10(2 + 57)$
$= 10 \times 59 = \boxed{590}$
If the first term of an A.P. is $5$, the last term is $15$ and the sum of first $n$ terms is $30$, then find the value of $n$.
Answer
Given: $a = 5$, last term $l = 15$, $S_n = 30$
Using $S_n = \dfrac{n}{2}(a + l)$:
$30 = \dfrac{n}{2}(5 + 15) = \dfrac{n}{2} \times 20$
$30 = 10n$
$\boxed{n = 3}$
Using $S_n = \dfrac{n}{2}(a + l)$:
$30 = \dfrac{n}{2}(5 + 15) = \dfrac{n}{2} \times 20$
$30 = 10n$
$\boxed{n = 3}$
Rohan repays his total loan of ₹1,18,000 by paying every month starting with the first instalment of ₹1,000. If he increases the instalment by ₹100 every month, what amount will be paid by him in the 30th instalment? What amount of loan has he paid after the 30th instalment?
Answer
First instalment $a = ₹1000$, common difference $d = ₹100$
30th instalment:
$a_{30} = a + (30 – 1)d = 1000 + 29 \times 100 = 1000 + 2900 = ₹3900$
Total paid after 30 instalments:
$S_{30} = \dfrac{30}{2}[2 \times 1000 + (30 – 1) \times 100]$
$= 15[2000 + 2900] = 15 \times 4900 = ₹73,500$
Rohan pays ₹3,900 in the 30th instalment and has paid ₹73,500 after 30 instalments.
30th instalment:
$a_{30} = a + (30 – 1)d = 1000 + 29 \times 100 = 1000 + 2900 = ₹3900$
Total paid after 30 instalments:
$S_{30} = \dfrac{30}{2}[2 \times 1000 + (30 – 1) \times 100]$
$= 15[2000 + 2900] = 15 \times 4900 = ₹73,500$
Rohan pays ₹3,900 in the 30th instalment and has paid ₹73,500 after 30 instalments.
The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.
Answer
Given: $S_{15} = 750$, $a = 15$
Step 1: Find $d$:
$S_{15} = \dfrac{15}{2}[2 \times 15 + 14d] = 750$
$\dfrac{15}{2}[30 + 14d] = 750$
$30 + 14d = 100 \Rightarrow 14d = 70 \Rightarrow d = 5$
Step 2: Find $a_{20}$:
$a_{20} = 15 + (20-1) \times 5 = 15 + 95 = \boxed{110}$
Step 1: Find $d$:
$S_{15} = \dfrac{15}{2}[2 \times 15 + 14d] = 750$
$\dfrac{15}{2}[30 + 14d] = 750$
$30 + 14d = 100 \Rightarrow 14d = 70 \Rightarrow d = 5$
Step 2: Find $a_{20}$:
$a_{20} = 15 + (20-1) \times 5 = 15 + 95 = \boxed{110}$
Show that the sum of all terms of an A.P. whose first term is $a$, the second term is $b$, and the last term is $c$ is equal to $\dfrac{(a + c)(b + c – 2a)}{2(b – a)}$.
Answer
First term $= a$, common difference $d = b – a$, last term $= c$
Step 1: Find $n$:
$a + (n-1)(b-a) = c$
$(n-1) = \dfrac{c-a}{b-a}$
$n = \dfrac{c – a + b – a}{b – a} = \dfrac{b + c – 2a}{b – a}$
Step 2: Find the sum:
$\text{Sum} = \dfrac{n}{2}(a + c) = \dfrac{b + c – 2a}{2(b – a)} \times (a + c) = \dfrac{(a+c)(b+c-2a)}{2(b-a)}$
Hence Proved.
Step 1: Find $n$:
$a + (n-1)(b-a) = c$
$(n-1) = \dfrac{c-a}{b-a}$
$n = \dfrac{c – a + b – a}{b – a} = \dfrac{b + c – 2a}{b – a}$
Step 2: Find the sum:
$\text{Sum} = \dfrac{n}{2}(a + c) = \dfrac{b + c – 2a}{2(b – a)} \times (a + c) = \dfrac{(a+c)(b+c-2a)}{2(b-a)}$
Hence Proved.
Solve the equation: $1 + 4 + 7 + 10 + \ldots + x = 287$.
Answer
$a = 1$, $d = 3$. Let the number of terms be $n$.
$S_n = \dfrac{n}{2}[2 + (n-1)3] = 287$
$n[2 + 3n – 3] = 574$
$3n^2 – n – 574 = 0$
$(n – 14)(3n + 41) = 0$
Since $n > 0$, we get $n = 14$.
The 14th term $= a + 13d = 1 + 39 = \boxed{x = 40}$
$S_n = \dfrac{n}{2}[2 + (n-1)3] = 287$
$n[2 + 3n – 3] = 574$
$3n^2 – n – 574 = 0$
$(n – 14)(3n + 41) = 0$
Since $n > 0$, we get $n = 14$.
The 14th term $= a + 13d = 1 + 39 = \boxed{x = 40}$
If in an A.P., the sum of first $m$ terms is $n$ and the sum of its first $n$ terms is $m$, then prove that the sum of its first $(m + n)$ terms is $-(m + n)$.
Answer
Let the first term be $a$ and common difference be $d$.
$S_m = n \Rightarrow \dfrac{m}{2}[2a + (m-1)d] = n$ …(i)
$S_n = m \Rightarrow \dfrac{n}{2}[2a + (n-1)d] = m$ …(ii)
Subtracting (ii) from (i) and simplifying:
$2(n – m) = (m – n)[2a + d(m + n – 1)]$
$-2 = 2a + d(m + n – 1)$
Now:
$S_{m+n} = \dfrac{m+n}{2}[2a + (m+n-1)d] = \dfrac{m+n}{2} \times (-2) = -(m + n)$
Hence Proved.
$S_m = n \Rightarrow \dfrac{m}{2}[2a + (m-1)d] = n$ …(i)
$S_n = m \Rightarrow \dfrac{n}{2}[2a + (n-1)d] = m$ …(ii)
Subtracting (ii) from (i) and simplifying:
$2(n – m) = (m – n)[2a + d(m + n – 1)]$
$-2 = 2a + d(m + n – 1)$
Now:
$S_{m+n} = \dfrac{m+n}{2}[2a + (m+n-1)d] = \dfrac{m+n}{2} \times (-2) = -(m + n)$
Hence Proved.
If the last term of an A.P. of 30 terms is 119 and the 8th term from the end (towards the first term) is 91, then find the common difference of the A.P. Hence, find the sum of all terms of the A.P.
Answer
Last term $l = 119$, total terms $= 30$, 8th term from end $= 91$
Step 1: Find $d$ from the end:
$a_8$ (from end) $= 119 + (8-1)d_{\text{end}} = 91$
$7d_{\text{end}} = -28 \Rightarrow d_{\text{end}} = -4$
Common difference from start $= -(-4) = \mathbf{4}$
Step 2: Find the first term:
$119 = a + 29 \times 4 \Rightarrow a = 119 – 116 = 3$
Step 3: Find the sum:
$S_{30} = \dfrac{30}{2}[2 \times 3 + 29 \times 4] = 15(6 + 116) = 15 \times 122 = \boxed{1830}$
Step 1: Find $d$ from the end:
$a_8$ (from end) $= 119 + (8-1)d_{\text{end}} = 91$
$7d_{\text{end}} = -28 \Rightarrow d_{\text{end}} = -4$
Common difference from start $= -(-4) = \mathbf{4}$
Step 2: Find the first term:
$119 = a + 29 \times 4 \Rightarrow a = 119 – 116 = 3$
Step 3: Find the sum:
$S_{30} = \dfrac{30}{2}[2 \times 3 + 29 \times 4] = 15(6 + 116) = 15 \times 122 = \boxed{1830}$
In an A.P., the sum of first $n$ terms is $S_n = \dfrac{n}{2}(3n + 5)$. Find the 25th term of the A.P.
Answer
$S_n = \dfrac{n}{2}(3n + 5)$ and $S_{n-1} = \dfrac{n-1}{2}(3n + 2)$
$a_n = S_n – S_{n-1}$
$= \dfrac{n}{2}(3n+5) – \dfrac{n-1}{2}(3n+2)$
$= \dfrac{3n^2 + 5n – (3n^2 – n – 2)}{2}$
$= \dfrac{6n + 2}{2} = 3n + 1$
$a_{25} = 3(25) + 1 = 75 + 1 = \boxed{76}$
$a_n = S_n – S_{n-1}$
$= \dfrac{n}{2}(3n+5) – \dfrac{n-1}{2}(3n+2)$
$= \dfrac{3n^2 + 5n – (3n^2 – n – 2)}{2}$
$= \dfrac{6n + 2}{2} = 3n + 1$
$a_{25} = 3(25) + 1 = 75 + 1 = \boxed{76}$
If the $n^{\text{th}}$ term of an A.P. is $(2n + 1)$, what is the sum of its first three terms?
Answer
$n^{\text{th}}$ term $a_n = 2n + 1$
$a_1 = 2(1) + 1 = 3$
$a_2 = 2(2) + 1 = 5$
$a_3 = 2(3) + 1 = 7$
Sum of first three terms $= 3 + 5 + 7 = \boxed{15}$
$a_1 = 2(1) + 1 = 3$
$a_2 = 2(2) + 1 = 5$
$a_3 = 2(3) + 1 = 7$
Sum of first three terms $= 3 + 5 + 7 = \boxed{15}$
How many natural numbers are there between 1 and 1000 which are divisible by 5 but not by 2?
Answer
Numbers between 1 and 1000 divisible by 5 but not 2: $5, 15, 25, 35, \ldots, 995$
(odd multiples of 5, common difference $= 10$)
Using $l = a + (n-1)d$:
$995 = 5 + (n-1) \times 10$
$990 = 10(n-1)$
$n – 1 = 99 \Rightarrow n = 100$
There are 100 such natural numbers.
(odd multiples of 5, common difference $= 10$)
Using $l = a + (n-1)d$:
$995 = 5 + (n-1) \times 10$
$990 = 10(n-1)$
$n – 1 = 99 \Rightarrow n = 100$
There are 100 such natural numbers.
Which term of the A.P.: $65, 61, 57, 53, \ldots$ is the first negative term?
Answer
$a = 65$, $d = -4$
For the first negative term, $a_n < 0$:
$65 + (n-1)(-4) < 0$
$69 – 4n < 0$
$n > \dfrac{69}{4} = 17.25$
The smallest integer greater than $17.25$ is $18$.
The 18th term is the first negative term.
For the first negative term, $a_n < 0$:
$65 + (n-1)(-4) < 0$
$69 – 4n < 0$
$n > \dfrac{69}{4} = 17.25$
The smallest integer greater than $17.25$ is $18$.
The 18th term is the first negative term.
The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last term to the product of the two middle terms is $7 : 15$. Find the numbers.
Answer
Let the four consecutive terms be $(a – 3d)$, $(a – d)$, $(a + d)$, $(a + 3d)$.
Condition 1 — Sum:
$(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32$
$4a = 32 \Rightarrow a = 8$
Condition 2 — Ratio:
$\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)} = \dfrac{7}{15}$
$\dfrac{a^2 – 9d^2}{a^2 – d^2} = \dfrac{7}{15}$
$15(64 – 9d^2) = 7(64 – d^2)$
$960 – 135d^2 = 448 – 7d^2$
$128d^2 = 512 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2$
The four numbers are $2, 6, 10, 14$ (or $14, 10, 6, 2$).
Condition 1 — Sum:
$(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32$
$4a = 32 \Rightarrow a = 8$
Condition 2 — Ratio:
$\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)} = \dfrac{7}{15}$
$\dfrac{a^2 – 9d^2}{a^2 – d^2} = \dfrac{7}{15}$
$15(64 – 9d^2) = 7(64 – d^2)$
$960 – 135d^2 = 448 – 7d^2$
$128d^2 = 512 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2$
The four numbers are $2, 6, 10, 14$ (or $14, 10, 6, 2$).
The ratio of the 11th term to the 18th term of an A.P. is $2 : 3$. Find the ratio of the 5th term to the 21st term. Also, find the ratio of the sum of first 5 terms to the sum of first 21 terms.
Answer
$\dfrac{a_{11}}{a_{18}} = \dfrac{a + 10d}{a + 17d} = \dfrac{2}{3}$
$3a + 30d = 2a + 34d \Rightarrow a = 4d$
Ratio of 5th to 21st term:
$\dfrac{a_5}{a_{21}} = \dfrac{a + 4d}{a + 20d} = \dfrac{4d + 4d}{4d + 20d} = \dfrac{8d}{24d} = \dfrac{1}{3}$
$\Rightarrow a_5 : a_{21} = \mathbf{1 : 3}$
Ratio of sum of first 5 to first 21 terms:
$\dfrac{S_5}{S_{21}} = \dfrac{\frac{5}{2}(2a + 4d)}{\frac{21}{2}(2a + 20d)} = \dfrac{5(8d + 4d)}{21(8d + 20d)} = \dfrac{5 \times 12d}{21 \times 28d} = \dfrac{60}{588} = \dfrac{5}{49}$
$\Rightarrow S_5 : S_{21} = \mathbf{5 : 49}$
$3a + 30d = 2a + 34d \Rightarrow a = 4d$
Ratio of 5th to 21st term:
$\dfrac{a_5}{a_{21}} = \dfrac{a + 4d}{a + 20d} = \dfrac{4d + 4d}{4d + 20d} = \dfrac{8d}{24d} = \dfrac{1}{3}$
$\Rightarrow a_5 : a_{21} = \mathbf{1 : 3}$
Ratio of sum of first 5 to first 21 terms:
$\dfrac{S_5}{S_{21}} = \dfrac{\frac{5}{2}(2a + 4d)}{\frac{21}{2}(2a + 20d)} = \dfrac{5(8d + 4d)}{21(8d + 20d)} = \dfrac{5 \times 12d}{21 \times 28d} = \dfrac{60}{588} = \dfrac{5}{49}$
$\Rightarrow S_5 : S_{21} = \mathbf{5 : 49}$
If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Answer
Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$:
$S_6 = 36 \Rightarrow 2a + 5d = 12$ …(i)
$S_{16} = 256 \Rightarrow 2a + 15d = 32$ …(ii)
Subtracting (i) from (ii): $10d = 20 \Rightarrow d = 2$
Substituting in (i): $2a + 10 = 12 \Rightarrow a = 1$
$S_{10} = \dfrac{10}{2}[2 \times 1 + 9 \times 2] = 5(2 + 18) = 5 \times 20 = \boxed{100}$
$S_6 = 36 \Rightarrow 2a + 5d = 12$ …(i)
$S_{16} = 256 \Rightarrow 2a + 15d = 32$ …(ii)
Subtracting (i) from (ii): $10d = 20 \Rightarrow d = 2$
Substituting in (i): $2a + 10 = 12 \Rightarrow a = 1$
$S_{10} = \dfrac{10}{2}[2 \times 1 + 9 \times 2] = 5(2 + 18) = 5 \times 20 = \boxed{100}$
The sum of first $q$ terms of an A.P. is $63q – 3q^2$. If its $p^{\text{th}}$ term is $-60$, find the value of $p$. Also, find the 11th term of this A.P.
Answer
$S_q = 63q – 3q^2$
$a_1 = S_1 = 63 – 3 = 60$
$a_2 = S_2 – S_1 = (126 – 12) – 60 = 54$
$d = 54 – 60 = -6$
Find $p$:
$a_p = a + (p-1)d = -60$
$60 + (p-1)(-6) = -60$
$(p-1)(-6) = -120 \Rightarrow p – 1 = 20 \Rightarrow p = 21$
Find $a_{11}$:
$a_{11} = 60 + 10 \times (-6) = 60 – 60 = \boxed{0}$
The value of $p$ is 21 and the 11th term is 0.
$a_1 = S_1 = 63 – 3 = 60$
$a_2 = S_2 – S_1 = (126 – 12) – 60 = 54$
$d = 54 – 60 = -6$
Find $p$:
$a_p = a + (p-1)d = -60$
$60 + (p-1)(-6) = -60$
$(p-1)(-6) = -120 \Rightarrow p – 1 = 20 \Rightarrow p = 21$
Find $a_{11}$:
$a_{11} = 60 + 10 \times (-6) = 60 – 60 = \boxed{0}$
The value of $p$ is 21 and the 11th term is 0.
Find the sum of integers between 100 and 200 which are:
(i) divisible by 9
(ii) not divisible by 9.
(i) divisible by 9
(ii) not divisible by 9.
Answer
(i) Integers between 100 and 200 divisible by 9:
Series: $108, 117, 126, \ldots, 198$ with $a = 108$, $d = 9$, $l = 198$
$198 = 108 + (n-1) \times 9 \Rightarrow n = 11$
$S_{11} = \dfrac{11}{2}(108 + 198) = \dfrac{11}{2} \times 306 = 11 \times 153 = \mathbf{1683}$
(ii) Integers between 100 and 200 not divisible by 9:
All integers: $101, 102, \ldots, 199$ with $a = 101$, $d = 1$, $l = 199$
$199 = 101 + (n-1) \Rightarrow n = 99$
$S_{99} = \dfrac{99}{2}(101 + 199) = \dfrac{99}{2} \times 300 = 14850$
Sum not divisible by 9 $= 14850 – 1683 = \mathbf{13167}$
Series: $108, 117, 126, \ldots, 198$ with $a = 108$, $d = 9$, $l = 198$
$198 = 108 + (n-1) \times 9 \Rightarrow n = 11$
$S_{11} = \dfrac{11}{2}(108 + 198) = \dfrac{11}{2} \times 306 = 11 \times 153 = \mathbf{1683}$
(ii) Integers between 100 and 200 not divisible by 9:
All integers: $101, 102, \ldots, 199$ with $a = 101$, $d = 1$, $l = 199$
$199 = 101 + (n-1) \Rightarrow n = 99$
$S_{99} = \dfrac{99}{2}(101 + 199) = \dfrac{99}{2} \times 300 = 14850$
Sum not divisible by 9 $= 14850 – 1683 = \mathbf{13167}$
Push-ups are a fast and effective exercise for building strength. Nitesh wants to participate in a push-up challenge. He can currently make 3000 push-ups in one hour but wants to achieve a target of 3900 push-ups in 1 hour. With each day of practice, he is able to make 5 more push-ups compared to the previous day. On the first day he makes 3000 push-ups.
(i) Form an A.P. representing the number of push-ups per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished.
(ii) Find the total number of push-ups performed by Nitesh up to the day his goal is achieved.
(i) Form an A.P. representing the number of push-ups per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished.
(ii) Find the total number of push-ups performed by Nitesh up to the day his goal is achieved.
Answer
$a_1 = 3000$, $d = 5$
(i) Minimum days of practice:
$a_n = 3000 + (n-1) \times 5 = 3900$
$(n-1) \times 5 = 900 \Rightarrow n – 1 = 180 \Rightarrow n = 181$
Nitesh needs to practice for a minimum of 181 days.
(ii) Total push-ups up to and including day 181:
$S_{181} = \dfrac{181}{2}(3000 + 3900) = \dfrac{181}{2} \times 6900 = 181 \times 3450 = \mathbf{624{,}450}$
Nitesh performs a total of 6,24,450 push-ups.
(i) Minimum days of practice:
$a_n = 3000 + (n-1) \times 5 = 3900$
$(n-1) \times 5 = 900 \Rightarrow n – 1 = 180 \Rightarrow n = 181$
Nitesh needs to practice for a minimum of 181 days.
(ii) Total push-ups up to and including day 181:
$S_{181} = \dfrac{181}{2}(3000 + 3900) = \dfrac{181}{2} \times 6900 = 181 \times 3450 = \mathbf{624{,}450}$
Nitesh performs a total of 6,24,450 push-ups.
The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement such that each succeeding circular row has 10 seats more than the previous one.
(i) If the first circular row has 30 seats, how many seats will be there in the 10th row?
(ii) For 1500 seats in the auditorium, how many rows need to be there?
(i) If the first circular row has 30 seats, how many seats will be there in the 10th row?
(ii) For 1500 seats in the auditorium, how many rows need to be there?
Answer
$a_1 = 30$, $d = 10$
(i) Seats in the 10th row:
$a_{10} = 30 + (10-1) \times 10 = 30 + 90 = \mathbf{120 \text{ seats}}$
(ii) Number of rows for 1500 seats:
$S_n = \dfrac{n}{2}[2(30) + (n-1)(10)] = \dfrac{n}{2}(60 + 10n – 10) = 5n(n + 5)$
Setting $S_n \geq 1500$:
$5n(n+5) \geq 1500 \Rightarrow n^2 + 5n – 300 \geq 0$
Using the quadratic formula:
$n = \dfrac{-5 + \sqrt{25 + 1200}}{2} = \dfrac{-5 + 35}{2} = 15$
15 rows are needed to accommodate at least 1500 seats.
(i) Seats in the 10th row:
$a_{10} = 30 + (10-1) \times 10 = 30 + 90 = \mathbf{120 \text{ seats}}$
(ii) Number of rows for 1500 seats:
$S_n = \dfrac{n}{2}[2(30) + (n-1)(10)] = \dfrac{n}{2}(60 + 10n – 10) = 5n(n + 5)$
Setting $S_n \geq 1500$:
$5n(n+5) \geq 1500 \Rightarrow n^2 + 5n – 300 \geq 0$
Using the quadratic formula:
$n = \dfrac{-5 + \sqrt{25 + 1200}}{2} = \dfrac{-5 + 35}{2} = 15$
15 rows are needed to accommodate at least 1500 seats.
Frequently Asked Questions
What does the Arithmetic Progressions chapter cover in CBSE Class 10 Maths? ⌄
The Arithmetic Progressions chapter covers the definition and general form of an A.P., finding the nth term using the formula $a_n = a + (n-1)d$, finding the sum of the first n terms using $S_n = \frac{n}{2}[2a + (n-1)d]$, and applying these concepts to real-life word problems. Students also learn to determine whether a given sequence is an A.P. and find missing terms.
How many marks does Arithmetic Progressions carry in the CBSE Class 10 board exam? ⌄
Arithmetic Progressions is part of the Algebra unit, which carries approximately 20 marks in the CBSE Class 10 Maths board paper. Questions from this chapter appear as 1-mark MCQs, 2-mark and 3-mark short answers, and 5-mark case study or long-answer questions — making it one of the highest-weightage chapters across all question types.
What are the most important topics in Class 10 Arithmetic Progressions for board exams? ⌄
The most frequently tested topics are: finding the nth term of an A.P., finding the sum of n terms, determining the number of terms when the first, last, and sum are given, and solving real-life problems on A.P. such as loan repayments, seating arrangements, and daily-practice scenarios. Questions involving the condition for three numbers to form an A.P. are also very common in MCQ sections.
What common mistakes do students make when solving Arithmetic Progressions questions? ⌄
A very common error is forgetting to subtract 1 in the formula $a_n = a + (n-1)d$, writing it incorrectly as $a + nd$. Students also confuse the two sum formulas — $S_n = \frac{n}{2}[2a + (n-1)d]$ and $S_n = \frac{n}{2}(a + l)$ — and apply the wrong one. In word problems, students sometimes misidentify the first term or common difference, leading to wrong answers despite correct working.
How does Angle Belearn help students score well in Arithmetic Progressions? ⌄
Angle Belearn’s CBSE specialists carefully curate chapter-wise question banks drawn from real board papers spanning 2015–2025, each paired with clear, step-by-step solutions. Students practising on Angle Belearn develop the habit of showing structured working — something that earns full marks in board exams. Regular practice with these verified questions builds both speed and accuracy so your child walks into the exam confident.
