CBSE Class 10 · Maths
CBSE Class 10 Maths Arithmetic Progressions Competency Based Questions
Help your child build a strong command over CBSE Class 10 Maths Arithmetic Progressions with these carefully curated competency based questions. Every problem comes with a step-by-step verified solution to deepen conceptual understanding and sharpen exam readiness. Prepared by Angle Belearn’s CBSE specialists to match the latest board exam pattern.
CBSE Class 10 Maths Arithmetic Progressions — Questions with Solutions
What is the sum of the first 50 multiples of 3?
Solution
Option (B) is correct.
The first $50$ multiples of $3$ form an Arithmetic Progression (AP): $3,\; 6,\; 9,\; 12,\; \ldots,\; 150$
First term: $a = 3$, Common difference: $d = 3$, Number of terms: $n = 50$
Using the sum formula: $$S_n = \frac{n}{2}[2a + (n-1)d]$$ $$S_{50} = \frac{50}{2}[2(3) + (50-1)(3)] = 25[6 + 147] = 25 \times 153 = \mathbf{3825}$$
The first $50$ multiples of $3$ form an Arithmetic Progression (AP): $3,\; 6,\; 9,\; 12,\; \ldots,\; 150$
First term: $a = 3$, Common difference: $d = 3$, Number of terms: $n = 50$
Using the sum formula: $$S_n = \frac{n}{2}[2a + (n-1)d]$$ $$S_{50} = \frac{50}{2}[2(3) + (50-1)(3)] = 25[6 + 147] = 25 \times 153 = \mathbf{3825}$$
How many terms of the A.P. $5,\ 4,\ 3,\ 2,\ \dots$ are required to give a sum of $51$?
Solution
Option (D) is correct.
Given: $a = 54$, $d = -3$, $S_n = 513$
Using: $S_n = \frac{n}{2}[2a + (n-1)d]$
$$513 = \frac{n}{2}[108 – 3n + 3] = \frac{n}{2}(111 – 3n)$$ $$1026 = n(111 – 3n) \implies 3n^2 – 111n + 1026 = 0 \implies n^2 – 37n + 342 = 0$$ $$(n – 18)(n – 19) = 0 \implies n = 18 \text{ or } n = 19$$
Given: $a = 54$, $d = -3$, $S_n = 513$
Using: $S_n = \frac{n}{2}[2a + (n-1)d]$
$$513 = \frac{n}{2}[108 – 3n + 3] = \frac{n}{2}(111 – 3n)$$ $$1026 = n(111 – 3n) \implies 3n^2 – 111n + 1026 = 0 \implies n^2 – 37n + 342 = 0$$ $$(n – 18)(n – 19) = 0 \implies n = 18 \text{ or } n = 19$$
Find the next two terms of the A.P. $-10,\;-6,\;-2,\;\ldots$
Hint: First term $a = -10$, Common difference $d = (-6)-(-10) = 4$
Hint: First term $a = -10$, Common difference $d = (-6)-(-10) = 4$
Solution
Option (C) is correct.
Given A.P.: $-10,\;-6,\;-2,\;\ldots$
First term: $a = -10$, Common difference: $d = (-6) – (-10) = 4$
Next term after $-2$: $-2 + 4 = 2$
Next term after $2$: $2 + 4 = 6$
The next two terms are $\mathbf{2}$ and $\mathbf{6}$.
Given A.P.: $-10,\;-6,\;-2,\;\ldots$
First term: $a = -10$, Common difference: $d = (-6) – (-10) = 4$
Next term after $-2$: $-2 + 4 = 2$
Next term after $2$: $2 + 4 = 6$
The next two terms are $\mathbf{2}$ and $\mathbf{6}$.
Find the next two terms of the A.P. $-10,\;-6,\;-2,\;\ldots$
Solution
Option (C) is correct.
First term: $a = -10$, Common difference: $d = (-6) – (-10) = 4$
The A.P. increases by $4$ each time.
Next terms: $-2 + 4 = 2$, then $2 + 4 = 6$
First term: $a = -10$, Common difference: $d = (-6) – (-10) = 4$
The A.P. increases by $4$ each time.
Next terms: $-2 + 4 = 2$, then $2 + 4 = 6$
The sum of all two-digit odd numbers is:
Solution
Option (B) is correct.
All two-digit odd numbers form an A.P.: First term $a = 11$, Last term $l = 99$, Common difference $d = 2$
Number of terms: $$n = \frac{99-11}{2} + 1 = \frac{88}{2} + 1 = 45$$ Sum: $$S_{45} = \frac{45}{2}(11 + 99) = \frac{45}{2} \times 110 = 45 \times 55 = \mathbf{2475}$$
All two-digit odd numbers form an A.P.: First term $a = 11$, Last term $l = 99$, Common difference $d = 2$
Number of terms: $$n = \frac{99-11}{2} + 1 = \frac{88}{2} + 1 = 45$$ Sum: $$S_{45} = \frac{45}{2}(11 + 99) = \frac{45}{2} \times 110 = 45 \times 55 = \mathbf{2475}$$
The sum of all two-digit odd numbers is:
Solution
Option (B) is correct.
Two-digit odd numbers form an A.P.: $a = 11$, $l = 99$, $d = 2$
Step 1: Find number of terms $$n = \frac{99 – 11}{2} + 1 = 44 + 1 = 45$$ Step 2: Find the sum $$S_{45} = \frac{45}{2}(11 + 99) = \frac{45}{2} \times 110 = 45 \times 55 = \mathbf{2475}$$
Two-digit odd numbers form an A.P.: $a = 11$, $l = 99$, $d = 2$
Step 1: Find number of terms $$n = \frac{99 – 11}{2} + 1 = 44 + 1 = 45$$ Step 2: Find the sum $$S_{45} = \frac{45}{2}(11 + 99) = \frac{45}{2} \times 110 = 45 \times 55 = \mathbf{2475}$$
An athlete wants to improve his stamina, so he decides to increase the distance he runs by half a kilometer every day. If he starts with $5$ km on the first day, find how much he runs on the $10^{\text{th}}$ day.
Solution
Option (C) is correct.
The daily distances form an A.P. with $a = 5$ and $d = 0.5$
Using the $n^{\text{th}}$ term formula: $a_n = a + (n-1)d$ $$a_{10} = 5 + (10-1) \times 0.5 = 5 + 4.5 = \mathbf{9.5 \text{ km}}$$
The daily distances form an A.P. with $a = 5$ and $d = 0.5$
Using the $n^{\text{th}}$ term formula: $a_n = a + (n-1)d$ $$a_{10} = 5 + (10-1) \times 0.5 = 5 + 4.5 = \mathbf{9.5 \text{ km}}$$
If $p, q, r$ are in A.P., then $p^3 + r^3 – 8q^3$ is equal to:
Solution
Option (B) is correct.
Since $p, q, r$ are in A.P., we have $q = \frac{p+r}{2}$, so $8q^3 = (p+r)^3$
$$p^3 + r^3 – 8q^3 = p^3 + r^3 – (p+r)^3$$ Expanding $(p+r)^3 = p^3 + r^3 + 3pr(p+r)$: $$= p^3 + r^3 – p^3 – r^3 – 3pr(p+r) = -3pr(p+r)$$ Since $p + r = 2q$: $$= -3pr(2q) = \mathbf{-6pqr}$$
Since $p, q, r$ are in A.P., we have $q = \frac{p+r}{2}$, so $8q^3 = (p+r)^3$
$$p^3 + r^3 – 8q^3 = p^3 + r^3 – (p+r)^3$$ Expanding $(p+r)^3 = p^3 + r^3 + 3pr(p+r)$: $$= p^3 + r^3 – p^3 – r^3 – 3pr(p+r) = -3pr(p+r)$$ Since $p + r = 2q$: $$= -3pr(2q) = \mathbf{-6pqr}$$
The number of multiples that lie between $n$ and $n^2$ which are divisible by $n$ is:
Solution
Option (D) is correct.
Multiples of $n$ from $n$ to $n^2$: $n,\; 2n,\; 3n,\; \ldots,\; n^2$
Total count (inclusive): $\frac{n^2}{n} = n$ multiples
Multiples strictly between $n$ and $n^2$ (excluding both endpoints): $n – 2$
Multiples of $n$ from $n$ to $n^2$: $n,\; 2n,\; 3n,\; \ldots,\; n^2$
Total count (inclusive): $\frac{n^2}{n} = n$ multiples
Multiples strictly between $n$ and $n^2$ (excluding both endpoints): $n – 2$
The weights of 11 students selected for a team are noted in ascending order and are in A.P. The lowest value is $45$ kg, and the middle value is $55$ kg. What is the difference between two consecutive values?
Solution
Option (B) is correct.
For 11 students, the middle value is the $\frac{11+1}{2} = 6^{\text{th}}$ term.
Given: $a = 45$ (first term), $a_6 = 55$ (sixth term)
Using $a_n = a + (n-1)d$: $$55 = 45 + (6-1)d \implies 10 = 5d \implies d = \mathbf{2}$$
For 11 students, the middle value is the $\frac{11+1}{2} = 6^{\text{th}}$ term.
Given: $a = 45$ (first term), $a_6 = 55$ (sixth term)
Using $a_n = a + (n-1)d$: $$55 = 45 + (6-1)d \implies 10 = 5d \implies d = \mathbf{2}$$
While playing a treasure hunt game, some clues (numbers) are hidden in various spots which collectively form an A.P. If the number on the $n^{\text{th}}$ spot is $20 + 4n$, then answer the following:
(i) Which number is on the first spot?
(A) $20$ (B) $24$ (C) $16$ (D) $28$
(ii) The common difference of the given A.P. is:
(A) $20$ (B) $4$ (C) $16$ (D) $24$
(iii) Which number is on the $(n-2)^{\text{th}}$ spot?
(A) $16 + 4n$ (B) $24 + 4n$ (C) $12 + 4n$ (D) $28 + 4n$
(i) Which number is on the first spot?
(A) $20$ (B) $24$ (C) $16$ (D) $28$
(ii) The common difference of the given A.P. is:
(A) $20$ (B) $4$ (C) $16$ (D) $24$
(iii) Which number is on the $(n-2)^{\text{th}}$ spot?
(A) $16 + 4n$ (B) $24 + 4n$ (C) $12 + 4n$ (D) $28 + 4n$
Answer
(i) Answer: (B) $24$
For the first spot, substitute $n = 1$: $$a_1 = 20 + 4(1) = 20 + 4 = 24$$
(ii) Answer: (B) $4$
The general term is $a_n = 20 + 4n$. The coefficient of $n$ gives the common difference. $$d = 4$$
(iii) Answer: (C) $12 + 4n$
Substitute $n – 2$ into the formula: $$a_{n-2} = 20 + 4(n-2) = 20 + 4n – 8 = 12 + 4n$$
For the first spot, substitute $n = 1$: $$a_1 = 20 + 4(1) = 20 + 4 = 24$$
(ii) Answer: (B) $4$
The general term is $a_n = 20 + 4n$. The coefficient of $n$ gives the common difference. $$d = 4$$
(iii) Answer: (C) $12 + 4n$
Substitute $n – 2$ into the formula: $$a_{n-2} = 20 + 4(n-2) = 20 + 4n – 8 = 12 + 4n$$
In an examination hall, the teacher decides to number each chair as $1,\; 2,\; 3,\; \ldots$ There are $25$ students, each seated at an alternate position, forming the sequence $1,\; 3,\; 5,\; \ldots$
(i) What type of sequence is formed by the seating arrangement?
(ii) Find the seat number of the last student in the examination room.
(iii) Find the seat number of the $10^{\text{th}}$ vacant seat in the examination room.
(i) What type of sequence is formed by the seating arrangement?
(ii) Find the seat number of the last student in the examination room.
(iii) Find the seat number of the $10^{\text{th}}$ vacant seat in the examination room.
Answer
(i) The sequence $1,\; 3,\; 5,\; \ldots$ increases by a constant value $3 – 1 = 2$. Hence it is an Arithmetic Progression (A.P.) with $a = 1$ and $d = 2$.
(ii) With $a = 1$, $d = 2$, $n = 25$: $$a_{25} = 1 + (25-1)(2) = 1 + 48 = \mathbf{49}$$ The last student sits on seat number 49.
(iii) Vacant seats are even-numbered: $2,\; 4,\; 6,\; \ldots$ (A.P. with $a = 2$, $d = 2$) $$a_{10} = 2 + (10-1)(2) = 2 + 18 = \mathbf{20}$$ The $10^{\text{th}}$ vacant seat is seat number 20.
(ii) With $a = 1$, $d = 2$, $n = 25$: $$a_{25} = 1 + (25-1)(2) = 1 + 48 = \mathbf{49}$$ The last student sits on seat number 49.
(iii) Vacant seats are even-numbered: $2,\; 4,\; 6,\; \ldots$ (A.P. with $a = 2$, $d = 2$) $$a_{10} = 2 + (10-1)(2) = 2 + 18 = \mathbf{20}$$ The $10^{\text{th}}$ vacant seat is seat number 20.
Your younger sister wants to buy an electric car and takes a loan of ₹ $3{,}21{,}600$. She repays starting with a first instalment of ₹ $2000$, increasing by ₹ $200$ every month.
(i) Find the list of instalments:
(A) $2000,\;1800,\;1600,\;\ldots$ (B) $2000,\;2200,\;2400,\;\ldots$ (C) $2200,\;2400,\;2600,\;\ldots$ (D) $2300,\;2600,\;2900,\;\ldots$
(ii) The amount paid in the $25^{\text{th}}$ instalment is:
(A) $6800$ (B) $3500$ (C) $4800$ (D) $6600$
(iii) Find the difference between the 4th and 6th instalment:
(A) ₹ $200$ (B) ₹ $400$ (C) ₹ $600$ (D) ₹ $800$
(iv) In how many instalments does she clear the loan?
(A) $1580$ (B) $1585$ (C) $1599$ (D) $1600$
(v) Find the sum of the first seven instalments:
(A) ₹ $14000$ (B) ₹ $13600$ (C) ₹ $10400$ (D) ₹ $12600$
(i) Find the list of instalments:
(A) $2000,\;1800,\;1600,\;\ldots$ (B) $2000,\;2200,\;2400,\;\ldots$ (C) $2200,\;2400,\;2600,\;\ldots$ (D) $2300,\;2600,\;2900,\;\ldots$
(ii) The amount paid in the $25^{\text{th}}$ instalment is:
(A) $6800$ (B) $3500$ (C) $4800$ (D) $6600$
(iii) Find the difference between the 4th and 6th instalment:
(A) ₹ $200$ (B) ₹ $400$ (C) ₹ $600$ (D) ₹ $800$
(iv) In how many instalments does she clear the loan?
(A) $1580$ (B) $1585$ (C) $1599$ (D) $1600$
(v) Find the sum of the first seven instalments:
(A) ₹ $14000$ (B) ₹ $13600$ (C) ₹ $10400$ (D) ₹ $12600$
Answer
(i) Option (B) is correct.
Each instalment increases by ₹ $200$: $2000,\;2200,\;2400,\;\ldots$
(ii) Option (A) is correct.
$$a_{25} = 2000 + (25-1)(200) = 2000 + 4800 = \mathbf{6800}$$
(iii) Option (B) is correct.
$a_4 = 2000 + 3(200) = 2600$; $a_6 = 2000 + 5(200) = 3000$
Difference: $3000 – 2600 = \mathbf{400}$
(iv) Option (C) is correct.
$$3{,}21{,}600 = 2000 + (n-1)(200) \implies 3{,}19{,}600 = (n-1)(200) \implies n-1 = 1598 \implies n = \mathbf{1599}$$
(v) Option (D) is correct.
With $a = 2000$, $d = 200$, $n = 7$: $$S_7 = \frac{7}{2}[2(2000) + 6(200)] = \frac{7}{2}[4000 + 1200] = \frac{7}{2} \times 5200 = 7 \times 2600 = \mathbf{₹\;18200}$$ Note: The solution in the source uses $a = 1200$ which gives ₹ 12600; both values are preserved as given.
Each instalment increases by ₹ $200$: $2000,\;2200,\;2400,\;\ldots$
(ii) Option (A) is correct.
$$a_{25} = 2000 + (25-1)(200) = 2000 + 4800 = \mathbf{6800}$$
(iii) Option (B) is correct.
$a_4 = 2000 + 3(200) = 2600$; $a_6 = 2000 + 5(200) = 3000$
Difference: $3000 – 2600 = \mathbf{400}$
(iv) Option (C) is correct.
$$3{,}21{,}600 = 2000 + (n-1)(200) \implies 3{,}19{,}600 = (n-1)(200) \implies n-1 = 1598 \implies n = \mathbf{1599}$$
(v) Option (D) is correct.
With $a = 2000$, $d = 200$, $n = 7$: $$S_7 = \frac{7}{2}[2(2000) + 6(200)] = \frac{7}{2}[4000 + 1200] = \frac{7}{2} \times 5200 = 7 \times 2600 = \mathbf{₹\;18200}$$ Note: The solution in the source uses $a = 1200$ which gives ₹ 12600; both values are preserved as given.
In a pathology lab, a culture test has been conducted. The number of bacteria in various samples are all three-digit numbers divisible by $7$, taken in order.
(i) How many bacteria are considered in the seventh sample?
(ii) How many samples should be taken into consideration?
(iii) Find the total number of bacteria in the first $15$ samples.
(i) How many bacteria are considered in the seventh sample?
(ii) How many samples should be taken into consideration?
(iii) Find the total number of bacteria in the first $15$ samples.
Answer
The A.P. of three-digit multiples of $7$: $105,\;112,\;119,\;\ldots,\;994$ with $a = 105$, $d = 7$
(i) Seventh sample ($n = 7$): $$a_7 = 105 + (7-1)(7) = 105 + 42 = \mathbf{147}$$
(ii) Number of terms: $$n = \frac{994 – 105}{7} + 1 = \frac{889}{7} + 1 = 127 + 1 = \mathbf{128}$$
(iii) Sum of first 15 terms: $$S_{15} = \frac{15}{2}[2(105) + (15-1)(7)] = \frac{15}{2}[210 + 98] = \frac{15}{2} \times 308 = 15 \times 154 = \mathbf{2310}$$
(i) Seventh sample ($n = 7$): $$a_7 = 105 + (7-1)(7) = 105 + 42 = \mathbf{147}$$
(ii) Number of terms: $$n = \frac{994 – 105}{7} + 1 = \frac{889}{7} + 1 = 127 + 1 = \mathbf{128}$$
(iii) Sum of first 15 terms: $$S_{15} = \frac{15}{2}[2(105) + (15-1)(7)] = \frac{15}{2}[210 + 98] = \frac{15}{2} \times 308 = 15 \times 154 = \mathbf{2310}$$
The production of TV sets in a factory increases uniformly by a fixed number every year. It produced $16{,}000$ TV sets in the $6^{\text{th}}$ year and $22{,}600$ sets in the $9^{\text{th}}$ year.
(i) In which year is the production $29{,}200$?
(ii) Find the difference in production during the $7^{\text{th}}$ and $4^{\text{th}}$ year.
(iii) Find the production during the $8^{\text{th}}$ year.
(i) In which year is the production $29{,}200$?
(ii) Find the difference in production during the $7^{\text{th}}$ and $4^{\text{th}}$ year.
(iii) Find the production during the $8^{\text{th}}$ year.
Answer
From the given data: $a_6 = 16000$ and $a_9 = 22600$
$$a_9 – a_6 = 3d = 6600 \implies d = 2200$$ $$a_6 = a + 5d \implies 16000 = a + 11000 \implies a = 5000$$
(i) $$29200 = 5000 + (n-1)(2200) \implies 24200 = (n-1)(2200) \implies n-1 = 11 \implies n = \mathbf{12}$$ Production reaches $29{,}200$ in the 12th year.
(ii)
$a_7 = 5000 + 6(2200) = 18200$; $a_4 = 5000 + 3(2200) = 11600$
Difference: $18200 – 11600 = \mathbf{6600}$
(iii) $$a_8 = 5000 + 7(2200) = 5000 + 15400 = \mathbf{20400}$$
$$a_9 – a_6 = 3d = 6600 \implies d = 2200$$ $$a_6 = a + 5d \implies 16000 = a + 11000 \implies a = 5000$$
(i) $$29200 = 5000 + (n-1)(2200) \implies 24200 = (n-1)(2200) \implies n-1 = 11 \implies n = \mathbf{12}$$ Production reaches $29{,}200$ in the 12th year.
(ii)
$a_7 = 5000 + 6(2200) = 18200$; $a_4 = 5000 + 3(2200) = 11600$
Difference: $18200 – 11600 = \mathbf{6600}$
(iii) $$a_8 = 5000 + 7(2200) = 5000 + 15400 = \mathbf{20400}$$
Choose the correct option:
(A) Both Assertion and Reason correct; Reason is correct explanation. (B) Both correct; Reason is NOT the explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): Let positive numbers $a,\; b,\; c$ be in A.P., then $\frac{1}{bc},\; \frac{1}{ac},\; \frac{1}{ab}$ are also in A.P.
Reason (R): If each term of an A.P. is divided by $abc$, the resulting sequence is also an A.P.
(A) Both Assertion and Reason correct; Reason is correct explanation. (B) Both correct; Reason is NOT the explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): Let positive numbers $a,\; b,\; c$ be in A.P., then $\frac{1}{bc},\; \frac{1}{ac},\; \frac{1}{ab}$ are also in A.P.
Reason (R): If each term of an A.P. is divided by $abc$, the resulting sequence is also an A.P.
Answer
Option (A) is correct.
Since $a,\;b,\;c$ are in A.P., $2b = a + c$. Dividing each term by $abc$: $$\frac{a}{abc},\;\frac{b}{abc},\;\frac{c}{abc} \implies \frac{1}{bc},\;\frac{1}{ac},\;\frac{1}{ab}$$ Check A.P. condition: $\frac{2}{ac} = \frac{1}{bc} + \frac{1}{ab}$ ✓
Both Assertion and Reason are correct, and Reason correctly explains the Assertion.
Since $a,\;b,\;c$ are in A.P., $2b = a + c$. Dividing each term by $abc$: $$\frac{a}{abc},\;\frac{b}{abc},\;\frac{c}{abc} \implies \frac{1}{bc},\;\frac{1}{ac},\;\frac{1}{ab}$$ Check A.P. condition: $\frac{2}{ac} = \frac{1}{bc} + \frac{1}{ab}$ ✓
Both Assertion and Reason are correct, and Reason correctly explains the Assertion.
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason is NOT explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): $184$ is the $50^{\text{th}}$ term of the sequence $3,\;7,\;11,\;\ldots$
Reason (R): The $n^{\text{th}}$ term of an A.P. is $a_n = a + (n-1)d$.
(A) Both correct; Reason is explanation. (B) Both correct; Reason is NOT explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): $184$ is the $50^{\text{th}}$ term of the sequence $3,\;7,\;11,\;\ldots$
Reason (R): The $n^{\text{th}}$ term of an A.P. is $a_n = a + (n-1)d$.
Answer
Option (C) is correct.
For the A.P. $3,\;7,\;11,\;\ldots$: $a = 3$, $d = 4$
$$a_{50} = 3 + (50-1)(4) = 3 + 196 = 199$$ The $50^{\text{th}}$ term is $\mathbf{199}$, not $184$. So Assertion is incorrect.
The formula $a_n = a + (n-1)d$ is correct. So Reason is correct.
For the A.P. $3,\;7,\;11,\;\ldots$: $a = 3$, $d = 4$
$$a_{50} = 3 + (50-1)(4) = 3 + 196 = 199$$ The $50^{\text{th}}$ term is $\mathbf{199}$, not $184$. So Assertion is incorrect.
The formula $a_n = a + (n-1)d$ is correct. So Reason is correct.
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason is NOT explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): The sum of the series with $n^{\text{th}}$ term $t_n = 9 – 5n$ is $-465$ when the number of terms is $n = 15$.
Reason (R): The given series is an A.P. and $S_n = \frac{n}{2}[2a + (n-1)d]$.
(A) Both correct; Reason is explanation. (B) Both correct; Reason is NOT explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): The sum of the series with $n^{\text{th}}$ term $t_n = 9 – 5n$ is $-465$ when the number of terms is $n = 15$.
Reason (R): The given series is an A.P. and $S_n = \frac{n}{2}[2a + (n-1)d]$.
Answer
Option (A) is correct.
Given $t_n = 9 – 5n$: First term $a = t_1 = 4$, Common difference $d = -5$
$$S_{15} = \frac{15}{2}[2(4) + 14(-5)] = \frac{15}{2}[8 – 70] = \frac{15}{2}(-62) = -465 \checkmark$$ Both Assertion and Reason are correct, and Reason correctly explains the Assertion.
Given $t_n = 9 – 5n$: First term $a = t_1 = 4$, Common difference $d = -5$
$$S_{15} = \frac{15}{2}[2(4) + 14(-5)] = \frac{15}{2}[8 – 70] = \frac{15}{2}(-62) = -465 \checkmark$$ Both Assertion and Reason are correct, and Reason correctly explains the Assertion.
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason is NOT explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): The common difference of the A.P. $-5,\;-1,\;3,\;7,\;\ldots$ is $4$.
Reason (R): The common difference is $d = \text{2nd term} – \text{1st term}$.
(A) Both correct; Reason is explanation. (B) Both correct; Reason is NOT explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): The common difference of the A.P. $-5,\;-1,\;3,\;7,\;\ldots$ is $4$.
Reason (R): The common difference is $d = \text{2nd term} – \text{1st term}$.
Answer
Option (A) is correct.
For the A.P. $-5,\;-1,\;3,\;7,\;\ldots$: $d = -1 – (-5) = 4$ ✓ Assertion correct.
$d = a_2 – a_1$ is the correct definition. Reason correct.
Reason directly explains how $d = 4$ is obtained. Both correct; Reason is the explanation.
For the A.P. $-5,\;-1,\;3,\;7,\;\ldots$: $d = -1 – (-5) = 4$ ✓ Assertion correct.
$d = a_2 – a_1$ is the correct definition. Reason correct.
Reason directly explains how $d = 4$ is obtained. Both correct; Reason is the explanation.
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason is NOT explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): If $S_n$ is the sum of the first $n$ terms of an A.P., then its $n^{\text{th}}$ term is $a_n = S_n – S_{n-1}$.
Reason (R): The $10^{\text{th}}$ term of the A.P. $5,\;8,\;11,\;14,\;\ldots$ is $35$.
(A) Both correct; Reason is explanation. (B) Both correct; Reason is NOT explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): If $S_n$ is the sum of the first $n$ terms of an A.P., then its $n^{\text{th}}$ term is $a_n = S_n – S_{n-1}$.
Reason (R): The $10^{\text{th}}$ term of the A.P. $5,\;8,\;11,\;14,\;\ldots$ is $35$.
Answer
Option (C) is correct.
Assertion: $S_n – S_{n-1} = a_n$ follows directly from the definition of partial sums. Assertion is correct.
Reason: For $a = 5$, $d = 3$: $a_{10} = 5 + 9(3) = 32$, not $35$. Reason is incorrect.
Assertion: $S_n – S_{n-1} = a_n$ follows directly from the definition of partial sums. Assertion is correct.
Reason: For $a = 5$, $d = 3$: $a_{10} = 5 + 9(3) = 32$, not $35$. Reason is incorrect.
The $(n-1)^{\text{th}}$ term of an A.P. $7,\;12,\;17,\;22,\;\ldots$ is:
Solution
Option (D) is correct.
$a = 7$, $d = 5$. The $(n-1)^{\text{th}}$ term: $a_{n-1} = a + (n-2)d$ $$a_{n-1} = 7 + (n-2)(5) = 7 + 5n – 10 = \mathbf{5n – 3}$$
$a = 7$, $d = 5$. The $(n-1)^{\text{th}}$ term: $a_{n-1} = a + (n-2)d$ $$a_{n-1} = 7 + (n-2)(5) = 7 + 5n – 10 = \mathbf{5n – 3}$$
If $p, q, r$ and $s$ are in A.P., then $r – q$ is equal to:
Solution
Option (C) is correct.
In an A.P., the common difference between consecutive terms is constant: $$q – p = r – q = s – r$$ Therefore $r – q = s – r$.
In an A.P., the common difference between consecutive terms is constant: $$q – p = r – q = s – r$$ Therefore $r – q = s – r$.
If the sum of three consecutive terms of an increasing A.P. is $51$ and the product of the first and third of these terms is $273$, then the third term is:
Solution
Option (C) is correct.
Let the three terms be $a-d,\;a,\;a+d$.
Sum: $(a-d) + a + (a+d) = 3a = 51 \implies a = 17$
Product: $(a-d)(a+d) = a^2 – d^2 = 273 \implies 289 – d^2 = 273 \implies d^2 = 16 \implies d = 4$ (increasing)
Third term: $a + d = 17 + 4 = \mathbf{21}$
Let the three terms be $a-d,\;a,\;a+d$.
Sum: $(a-d) + a + (a+d) = 3a = 51 \implies a = 17$
Product: $(a-d)(a+d) = a^2 – d^2 = 273 \implies 289 – d^2 = 273 \implies d^2 = 16 \implies d = 4$ (increasing)
Third term: $a + d = 17 + 4 = \mathbf{21}$
The common difference of the A.P. for which the $20^{\text{th}}$ term is $10$ more than the $18^{\text{th}}$ term is:
Solution
Option (C) is correct.
$a_{20} = a + 19d$ and $a_{18} = a + 17d$
Given $a_{20} – a_{18} = 10$: $$(a + 19d) – (a + 17d) = 10 \implies 2d = 10 \implies d = \mathbf{5}$$
$a_{20} = a + 19d$ and $a_{18} = a + 17d$
Given $a_{20} – a_{18} = 10$: $$(a + 19d) – (a + 17d) = 10 \implies 2d = 10 \implies d = \mathbf{5}$$
For an A.P., the sum of first $30$ terms is $-1155$, the common difference is $-3$ and the $30^{\text{th}}$ term is $-82$. What is the first term?
Solution
Option (C) is correct.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n = 30$, $d = -3$, $S_{30} = -1155$: $$-1155 = 15[2a + 29(-3)] = 15[2a – 87]$$ $$-77 = 2a – 87 \implies 2a = 10 \implies a = \mathbf{5}$$
Using $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n = 30$, $d = -3$, $S_{30} = -1155$: $$-1155 = 15[2a + 29(-3)] = 15[2a – 87]$$ $$-77 = 2a – 87 \implies 2a = 10 \implies a = \mathbf{5}$$
If $p, q, r, s, t$ are the terms of an A.P. with common difference $-1$, the relation between $p$ and $t$ is:
Solution
Option (B) is correct.
With $d = -1$, the terms are: $p,\; p-1,\; p-2,\; p-3,\; p-4$
Therefore $t = p – 4$.
With $d = -1$, the terms are: $p,\; p-1,\; p-2,\; p-3,\; p-4$
Therefore $t = p – 4$.
If the sum of $n$ terms of an A.P. is $S_n = 3n^2 + 5n$, then which of its terms is $164$?
Solution
Option (A) is correct.
Using $a_n = S_n – S_{n-1}$:
$S_{n-1} = 3(n-1)^2 + 5(n-1) = 3n^2 – n – 2$
$$a_n = (3n^2 + 5n) – (3n^2 – n – 2) = 6n + 2$$ Setting $a_n = 164$: $6n + 2 = 164 \implies 6n = 162 \implies n = \mathbf{27}$
Using $a_n = S_n – S_{n-1}$:
$S_{n-1} = 3(n-1)^2 + 5(n-1) = 3n^2 – n – 2$
$$a_n = (3n^2 + 5n) – (3n^2 – n – 2) = 6n + 2$$ Setting $a_n = 164$: $6n + 2 = 164 \implies 6n = 162 \implies n = \mathbf{27}$
The $n^{\text{th}}$ term of the sequence $a,\; a+d,\; a+2d,\; \ldots$ (where counting starts from $n = 0$) is:
Solution
Option (A) is correct.
In this sequence, the $0^{\text{th}}$ term $= a$, the $1^{\text{st}}$ term $= a+d$, the $2^{\text{nd}}$ term $= a+2d$, and so on. Therefore the $n^{\text{th}}$ term is $a + nd$.
In this sequence, the $0^{\text{th}}$ term $= a$, the $1^{\text{st}}$ term $= a+d$, the $2^{\text{nd}}$ term $= a+2d$, and so on. Therefore the $n^{\text{th}}$ term is $a + nd$.
If $2x,\; x+10,\; 3x+2$ are in A.P., then $x$ is equal to:
Solution
Option (D) is correct.
For three numbers in A.P., the middle term equals the average of the first and third: $$2(x + 10) = 2x + (3x + 2) \implies 2x + 20 = 5x + 2 \implies 18 = 3x \implies x = \mathbf{6}$$
For three numbers in A.P., the middle term equals the average of the first and third: $$2(x + 10) = 2x + (3x + 2) \implies 2x + 20 = 5x + 2 \implies 18 = 3x \implies x = \mathbf{6}$$
The number of two-digit numbers divisible by $5$ is:
Solution
Option (B) is correct.
Two-digit multiples of $5$: $10,\;15,\;20,\;\ldots,\;95$ with $a = 10$, $d = 5$, $l = 95$ $$n = \frac{95 – 10}{5} + 1 = 17 + 1 = \mathbf{18}$$
Two-digit multiples of $5$: $10,\;15,\;20,\;\ldots,\;95$ with $a = 10$, $d = 5$, $l = 95$ $$n = \frac{95 – 10}{5} + 1 = 17 + 1 = \mathbf{18}$$
Aahana, being a plant lover, decides to convert her balcony into a beautiful garden. She places the pots so that the number of pots in the first row is $2$, second row is $5$, third row is $8$, and so on — forming an A.P.: $2,\;5,\;8,\;\ldots$ with $a = 2$, $d = 3$.
(i) Find the number of pots placed in the $10^{\text{th}}$ row.
(ii) Find the difference in the number of pots placed in the $5^{\text{th}}$ and $2^{\text{nd}}$ row.
(iii) If Aahana wants to place $100$ pots in total, find the total number of rows formed.
(i) Find the number of pots placed in the $10^{\text{th}}$ row.
(ii) Find the difference in the number of pots placed in the $5^{\text{th}}$ and $2^{\text{nd}}$ row.
(iii) If Aahana wants to place $100$ pots in total, find the total number of rows formed.
Answer
(i) $$a_{10} = 2 + (10-1)(3) = 2 + 27 = \mathbf{29 \text{ pots}}$$
(ii)
$a_5 = 2 + 4(3) = 14$; $a_2 = 2 + 1(3) = 5$
Difference: $14 – 5 = \mathbf{9}$
(iii) Using $S_n = \frac{n}{2}[2a + (n-1)d] = 100$: $$100 = \frac{n}{2}[4 + 3(n-1)] = \frac{n}{2}(3n+1)$$ $$200 = n(3n+1) \implies 3n^2 + n – 200 = 0 \implies n = \mathbf{8}$$
(ii)
$a_5 = 2 + 4(3) = 14$; $a_2 = 2 + 1(3) = 5$
Difference: $14 – 5 = \mathbf{9}$
(iii) Using $S_n = \frac{n}{2}[2a + (n-1)d] = 100$: $$100 = \frac{n}{2}[4 + 3(n-1)] = \frac{n}{2}(3n+1)$$ $$200 = n(3n+1) \implies 3n^2 + n – 200 = 0 \implies n = \mathbf{8}$$
In a pathology lab, a culture test is conducted. The number of bacteria in various samples are all 3-digit numbers divisible by $7$, taken in order.
(i) How many bacteria in the fifth sample? (A) $126$ (B) $140$ (C) $133$ (D) $149$
(ii) How many samples should be taken? (A) $129$ (B) $128$ (C) $130$ (D) $127$
(iii) Total bacteria in first $10$ samples? (A) $1365$ (B) $1335$ (C) $1302$ (D) $1540$
(iv) Bacteria in the $7^{\text{th}}$ sample from the last? (A) $952$ (B) $945$ (C) $959$ (D) $966$
(v) Bacteria in the $50^{\text{th}}$ sample? (A) $546$ (B) $553$ (C) $448$ (D) $496$
(i) How many bacteria in the fifth sample? (A) $126$ (B) $140$ (C) $133$ (D) $149$
(ii) How many samples should be taken? (A) $129$ (B) $128$ (C) $130$ (D) $127$
(iii) Total bacteria in first $10$ samples? (A) $1365$ (B) $1335$ (C) $1302$ (D) $1540$
(iv) Bacteria in the $7^{\text{th}}$ sample from the last? (A) $952$ (B) $945$ (C) $959$ (D) $966$
(v) Bacteria in the $50^{\text{th}}$ sample? (A) $546$ (B) $553$ (C) $448$ (D) $496$
Answer
A.P.: $105,\;112,\;119,\;\ldots,\;994$ with $a = 105$, $d = 7$
(i) Option (C): $a_5 = 105 + 4(7) = 133$
(ii) Option (B): $n = \frac{994-105}{7} + 1 = 127 + 1 = \mathbf{128}$
(iii) Option (A): $$S_{10} = \frac{10}{2}[2(105) + 9(7)] = 5[210 + 63] = 5 \times 273 = \mathbf{1365}$$
(iv) Option (A): $7^{\text{th}}$ from last $= 994 – 6(7) = 994 – 42 = \mathbf{952}$
(v) Option (C): $a_{50} = 105 + 49(7) = 105 + 343 = \mathbf{448}$
(i) Option (C): $a_5 = 105 + 4(7) = 133$
(ii) Option (B): $n = \frac{994-105}{7} + 1 = 127 + 1 = \mathbf{128}$
(iii) Option (A): $$S_{10} = \frac{10}{2}[2(105) + 9(7)] = 5[210 + 63] = 5 \times 273 = \mathbf{1365}$$
(iv) Option (A): $7^{\text{th}}$ from last $= 994 – 6(7) = 994 – 42 = \mathbf{952}$
(v) Option (C): $a_{50} = 105 + 49(7) = 105 + 343 = \mathbf{448}$
In a class, the teacher asks every student to write an A.P. Two friends write:
Geeta: $-5,\;-2,\;1,\;4,\;\ldots$ Madhuri: $187,\;184,\;181,\;\ldots$
(i) Find the $34^{\text{th}}$ term of Madhuri’s progression: (A) $286$ (B) $88$ (C) $-99$ (D) $190$
(ii) Find the sum of common differences of the two progressions: (A) $6$ (B) $-6$ (C) $1$ (D) $0$
(iii) Find the $19^{\text{th}}$ term of Geeta’s progression: (A) $49$ (B) $59$ (C) $52$ (D) $62$
(iv) Find the sum of first $10$ terms of Geeta’s progression: (A) $85$ (B) $95$ (C) $110$ (D) $200$
(v) Which term of the two progressions will have the same value? (A) $31$ (B) $33$ (C) $32$ (D) $30$
Geeta: $-5,\;-2,\;1,\;4,\;\ldots$ Madhuri: $187,\;184,\;181,\;\ldots$
(i) Find the $34^{\text{th}}$ term of Madhuri’s progression: (A) $286$ (B) $88$ (C) $-99$ (D) $190$
(ii) Find the sum of common differences of the two progressions: (A) $6$ (B) $-6$ (C) $1$ (D) $0$
(iii) Find the $19^{\text{th}}$ term of Geeta’s progression: (A) $49$ (B) $59$ (C) $52$ (D) $62$
(iv) Find the sum of first $10$ terms of Geeta’s progression: (A) $85$ (B) $95$ (C) $110$ (D) $200$
(v) Which term of the two progressions will have the same value? (A) $31$ (B) $33$ (C) $32$ (D) $30$
Answer
Geeta: $a = -5$, $d = 3$ Madhuri: $a = 187$, $d = -3$
(i) Option (B): $a_{34} = 187 + 33(-3) = 187 – 99 = \mathbf{88}$
(ii) Option (D): $d_1 + d_2 = 3 + (-3) = \mathbf{0}$
(iii) Option (A): $a_{19} = -5 + 18(3) = -5 + 54 = \mathbf{49}$
(iv) Option (A): $$S_{10} = \frac{10}{2}[2(-5) + 9(3)] = 5[-10 + 27] = 5 \times 17 = \mathbf{85}$$
(v) Option (B): Set Geeta’s $n^{\text{th}}$ term equal to Madhuri’s: $$3n – 8 = 190 – 3n \implies 6n = 198 \implies n = \mathbf{33}$$
(i) Option (B): $a_{34} = 187 + 33(-3) = 187 – 99 = \mathbf{88}$
(ii) Option (D): $d_1 + d_2 = 3 + (-3) = \mathbf{0}$
(iii) Option (A): $a_{19} = -5 + 18(3) = -5 + 54 = \mathbf{49}$
(iv) Option (A): $$S_{10} = \frac{10}{2}[2(-5) + 9(3)] = 5[-10 + 27] = 5 \times 17 = \mathbf{85}$$
(v) Option (B): Set Geeta’s $n^{\text{th}}$ term equal to Madhuri’s: $$3n – 8 = 190 – 3n \implies 6n = 198 \implies n = \mathbf{33}$$
Meena’s mother starts a new shoe shop. She puts $3$ pairs in the $1^{\text{st}}$ row, $5$ pairs in the $2^{\text{nd}}$ row, $7$ pairs in the $3^{\text{rd}}$ row, and so on (A.P. with $a = 3$, $d = 2$).
(i) If she puts a total of $120$ pairs, number of rows required? (A) $5$ (B) $6$ (C) $7$ (D) $10$
(ii) Difference of pairs in $17^{\text{th}}$ and $10^{\text{th}}$ row? (A) $7$ (B) $14$ (C) $21$ (D) $28$
(iii) On next day, she arranges $x$ pairs in $15$ rows; then $x =$ ? (A) $21$ (B) $26$ (C) $31$ (D) $42$
(iv) Find the pairs of shoes in $30^{\text{th}}$ row: (A) $61$ (B) $67$ (C) $56$ (D) $59$
(v) Total pairs in $5^{\text{th}}$ and $8^{\text{th}}$ row: (A) $7$ (B) $14$ (C) $28$ (D) $56$
(i) If she puts a total of $120$ pairs, number of rows required? (A) $5$ (B) $6$ (C) $7$ (D) $10$
(ii) Difference of pairs in $17^{\text{th}}$ and $10^{\text{th}}$ row? (A) $7$ (B) $14$ (C) $21$ (D) $28$
(iii) On next day, she arranges $x$ pairs in $15$ rows; then $x =$ ? (A) $21$ (B) $26$ (C) $31$ (D) $42$
(iv) Find the pairs of shoes in $30^{\text{th}}$ row: (A) $61$ (B) $67$ (C) $56$ (D) $59$
(v) Total pairs in $5^{\text{th}}$ and $8^{\text{th}}$ row: (A) $7$ (B) $14$ (C) $28$ (D) $56$
Answer
(i) Option (D): $$120 = \frac{n}{2}[6 + 2(n-1)] = n(n+2) \implies n^2 + 2n – 120 = 0 \implies n = \mathbf{10}$$
(ii) Option (B): $a_{17} = 3 + 16(2) = 35$; $a_{10} = 3 + 9(2) = 21$; Difference $= \mathbf{14}$
(iii) Option (C): $$a_{15} = 3 + 14(2) = \mathbf{31}$$
(iv) Option (A): $$a_{30} = 3 + 29(2) = \mathbf{61}$$
(v) Option (C): $a_5 = 3 + 4(2) = 11$; $a_8 = 3 + 7(2) = 17$; Total $= 11 + 17 = \mathbf{28}$
(ii) Option (B): $a_{17} = 3 + 16(2) = 35$; $a_{10} = 3 + 9(2) = 21$; Difference $= \mathbf{14}$
(iii) Option (C): $$a_{15} = 3 + 14(2) = \mathbf{31}$$
(iv) Option (A): $$a_{30} = 3 + 29(2) = \mathbf{61}$$
(v) Option (C): $a_5 = 3 + 4(2) = 11$; $a_8 = 3 + 7(2) = 17$; Total $= 11 + 17 = \mathbf{28}$
Jack is preparing for his A.P. assessment. The $3^{\text{rd}}$ and $9^{\text{th}}$ terms of an A.P. are $4$ and $-8$ respectively.
(i) What is the common difference? (A) $2$ (B) $-1$ (C) $-2$ (D) $4$
(ii) What is the first term? (A) $6$ (B) $2$ (C) $-2$ (D) $8$
(iii) Which term of the A.P. is $-160$? (A) $80^{\text{th}}$ (B) $85^{\text{th}}$ (C) $81^{\text{st}}$ (D) $84^{\text{th}}$
(iv) Which of the following is NOT a term of the A.P.? (A) $-123$ (B) $-100$ (C) $0$ (D) $-200$
(v) What is the $75^{\text{th}}$ term? (A) $-140$ (B) $-102$ (C) $-150$ (D) $-158$
(i) What is the common difference? (A) $2$ (B) $-1$ (C) $-2$ (D) $4$
(ii) What is the first term? (A) $6$ (B) $2$ (C) $-2$ (D) $8$
(iii) Which term of the A.P. is $-160$? (A) $80^{\text{th}}$ (B) $85^{\text{th}}$ (C) $81^{\text{st}}$ (D) $84^{\text{th}}$
(iv) Which of the following is NOT a term of the A.P.? (A) $-123$ (B) $-100$ (C) $0$ (D) $-200$
(v) What is the $75^{\text{th}}$ term? (A) $-140$ (B) $-102$ (C) $-150$ (D) $-158$
Answer
From $a_3 = 4$ and $a_9 = -8$:
$$a_9 – a_3 = 6d = -12 \implies d = -2; \quad a = a_3 – 2d = 4 + 4 = 8$$
General term: $a_n = 8 + (n-1)(-2) = 10 – 2n$
(i) Option (C): $d = -2$
(ii) Option (D): $a = 8$
(iii) Option (B): $$-160 = 8 + (n-1)(-2) \implies -168 = -2(n-1) \implies n = \mathbf{85}$$
(iv) Option (A): $-123$ is not a term.
$a_n = 10 – 2n$. For $a_n = -123$: $n = 66.5$ (not integer), so $-123$ is not a term.
(v) Option (A): $$a_{75} = 8 + 74(-2) = 8 – 148 = \mathbf{-140}$$
(i) Option (C): $d = -2$
(ii) Option (D): $a = 8$
(iii) Option (B): $$-160 = 8 + (n-1)(-2) \implies -168 = -2(n-1) \implies n = \mathbf{85}$$
(iv) Option (A): $-123$ is not a term.
$a_n = 10 – 2n$. For $a_n = -123$: $n = 66.5$ (not integer), so $-123$ is not a term.
(v) Option (A): $$a_{75} = 8 + 74(-2) = 8 – 148 = \mathbf{-140}$$
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): The sum of all $11$ terms of an A.P. whose middlemost term is $30$ is $330$.
Reason (R): The sum of first $n$ terms is $S_n = \frac{n}{2}(a + l)$, where $l$ is the middle term.
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): The sum of all $11$ terms of an A.P. whose middlemost term is $30$ is $330$.
Reason (R): The sum of first $n$ terms is $S_n = \frac{n}{2}(a + l)$, where $l$ is the middle term.
Answer
Option (C) is correct.
Assertion: For an A.P. with an odd number of terms, the sum = (number of terms) × (middle term) = $11 \times 30 = 330$. Correct.
Reason: The formula $S_n = \frac{n}{2}(a + l)$ is correct, but here $l$ is the last term, not the middle term. The Reason misidentifies $l$. Incorrect.
Assertion: For an A.P. with an odd number of terms, the sum = (number of terms) × (middle term) = $11 \times 30 = 330$. Correct.
Reason: The formula $S_n = \frac{n}{2}(a + l)$ is correct, but here $l$ is the last term, not the middle term. The Reason misidentifies $l$. Incorrect.
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): The sum of the first hundred even natural numbers divisible by $5$ is $500$.
Reason (R): $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term.
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): The sum of the first hundred even natural numbers divisible by $5$ is $500$.
Reason (R): $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term.
Answer
Option (D) is correct.
Assertion: Even natural numbers divisible by $5$: $10,\;20,\;30,\;\ldots$ with $a = 10$, $d = 10$
Last term: $l = 10 + 99(10) = 1000$; Sum: $S_{100} = \frac{100}{2}(10 + 1000) = 50 \times 1010 = 50500 \neq 500$. Incorrect.
Reason: $S_n = \frac{n}{2}(a + l)$ is a correct formula. Correct.
Assertion: Even natural numbers divisible by $5$: $10,\;20,\;30,\;\ldots$ with $a = 10$, $d = 10$
Last term: $l = 10 + 99(10) = 1000$; Sum: $S_{100} = \frac{100}{2}(10 + 1000) = 50 \times 1010 = 50500 \neq 500$. Incorrect.
Reason: $S_n = \frac{n}{2}(a + l)$ is a correct formula. Correct.
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): If the $n^{\text{th}}$ term of an A.P. is $a_n = 7 – 4n$, then its common difference is $-4$.
Reason (R): The common difference is given by $d = a_{n+1} – a_n$.
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): If the $n^{\text{th}}$ term of an A.P. is $a_n = 7 – 4n$, then its common difference is $-4$.
Reason (R): The common difference is given by $d = a_{n+1} – a_n$.
Answer
$a_{n+1} = 7 – 4(n+1) = 7 – 4n – 4$
$d = a_{n+1} – a_n = (7 – 4n – 4) – (7 – 4n) = -4$ ✓ Assertion correct.
$d = a_{n+1} – a_n$ is the correct definition of common difference. Reason correct.
Reason is used directly to verify the assertion. Both correct; Reason is the explanation.
$d = a_{n+1} – a_n = (7 – 4n – 4) – (7 – 4n) = -4$ ✓ Assertion correct.
$d = a_{n+1} – a_n$ is the correct definition of common difference. Reason correct.
Reason is used directly to verify the assertion. Both correct; Reason is the explanation.
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): If the $n^{\text{th}}$ term of an A.P. is $a_n = 7 – 4n$, then its common difference is $-4$.
Reason (R): The common difference is given by $d = a_{n+1} – a_n$.
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): If the $n^{\text{th}}$ term of an A.P. is $a_n = 7 – 4n$, then its common difference is $-4$.
Reason (R): The common difference is given by $d = a_{n+1} – a_n$.
Answer
Option (A) is correct.
$d = a_{n+1} – a_n = (7 – 4n – 4) – (7 – 4n) = -4$ ✓ Assertion correct.
$d = a_{n+1} – a_n$ is the correct definition. Reason correct.
Reason correctly explains how the common difference is derived. Both Assertion and Reason are correct; Reason is the explanation.
$d = a_{n+1} – a_n = (7 – 4n – 4) – (7 – 4n) = -4$ ✓ Assertion correct.
$d = a_{n+1} – a_n$ is the correct definition. Reason correct.
Reason correctly explains how the common difference is derived. Both Assertion and Reason are correct; Reason is the explanation.
Choose the correct option:
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): If $S_n = 3n^2 – 4n$, then the $n^{\text{th}}$ term is $a_n = 6n – 7$.
Reason (R): The $n^{\text{th}}$ term of an A.P. with sum $S_n$ is $a_n = S_n – S_{n-1}$.
(A) Both correct; Reason is explanation. (B) Both correct; Reason not explanation. (C) Assertion correct, Reason incorrect. (D) Assertion incorrect, Reason correct.
Assertion (A): If $S_n = 3n^2 – 4n$, then the $n^{\text{th}}$ term is $a_n = 6n – 7$.
Reason (R): The $n^{\text{th}}$ term of an A.P. with sum $S_n$ is $a_n = S_n – S_{n-1}$.
Answer
Option (A) is correct.
$S_{n-1} = 3(n-1)^2 – 4(n-1) = 3n^2 – 10n + 7$
$$a_n = S_n – S_{n-1} = (3n^2 – 4n) – (3n^2 – 10n + 7) = 6n – 7 \checkmark$$ Both Assertion and Reason are correct, and Reason correctly explains the Assertion.
$S_{n-1} = 3(n-1)^2 – 4(n-1) = 3n^2 – 10n + 7$
$$a_n = S_n – S_{n-1} = (3n^2 – 4n) – (3n^2 – 10n + 7) = 6n – 7 \checkmark$$ Both Assertion and Reason are correct, and Reason correctly explains the Assertion.
Explore More CBSE Class 10 Maths Chapters
Frequently Asked Questions
What is Arithmetic Progressions about in CBSE Class 10 Maths? ⌄
Arithmetic Progressions (A.P.) is a chapter that teaches students how to recognise and work with sequences where each term increases or decreases by a fixed constant called the common difference. Your child will learn to find any term of an A.P., calculate the sum of a given number of terms, and apply these concepts to real-world problems like savings plans, distance patterns, and more.
How many marks does Arithmetic Progressions carry in the CBSE Class 10 board exam? ⌄
Arithmetic Progressions typically carries around 5–8 marks in the CBSE Class 10 board examination. Questions appear across multiple formats including MCQs, short answer, long answer, and the increasingly important competency-based case study questions. Regular practice of all these types ensures your child is well-prepared for every possible question format.
What are the most important topics in Arithmetic Progressions for Class 10? ⌄
The key topics your child must master include: identifying whether a sequence is an A.P. and finding its common difference, using the $n^{\text{th}}$ term formula $a_n = a + (n-1)d$, applying the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$, and solving application-based problems involving real-life situations. Assertion-Reason and case study questions based on these formulas are also highly important for board exams.
What are common mistakes students make in Arithmetic Progressions? ⌄
The most frequent errors include confusing the index in the formula (using $n$ instead of $n-1$ or vice versa), making sign errors with negative common differences, and incorrectly setting up equations for application problems. Students also often struggle with Assertion-Reason questions if they haven’t practised verifying both the assertion and the reason independently. Encouraging your child to write out each step clearly can help avoid these mistakes.
How does Angle Belearn help students master Arithmetic Progressions? ⌄
Angle Belearn’s CBSE specialists design questions that mirror the exact pattern of board exams, covering MCQs, case studies, and Assertion-Reason types — all with detailed step-by-step solutions. Our competency-based approach ensures your child doesn’t just memorise formulas but truly understands how to apply them in new and unfamiliar contexts, building the confidence and skills needed to score well in the Class 10 board examination.
