Board Previous Year Questions (Board PYQs)

How to Approach Board PYQs

This section collects the types of AP questions that repeatedly appear in CBSE and State Board examinations, with full step-by-step solutions. Working through these 26 problems is the best way to predict your exam.

What examiners love in this chapter:

1. nth term and 'which term' problems (2–3 marks).
2. Sum of $n$ terms, and finding $n$ from a given sum (3 marks).
3. Find $a$ and $d$ from two given terms or sums (3 marks).
4. Word problems — savings, seats, logs, production (3–4 marks).

Tag note: Questions are tagged by exam (e.g. [CBSE Board]); where the exact year is unconfirmed, only the exam name is given.

Exam Tip: Quote the formula, substitute carefully (mind the $(n-1)$), and state word-problem answers in words with units.

Solved Previous Year Questions

PYQ 1: Find the nth term (2 marks)

Find the 11th term of the AP $-3, -\dfrac{1}{2}, 2, \dots$ [CBSE Board]

Solution:

1. $a = -3$, $d = -\dfrac{1}{2} - (-3) = \dfrac{5}{2}$.
2. $a_{11} = -3 + 10\left(\dfrac{5}{2}\right) = -3 + 25 = 22$.

Final Answer: 22.

Takeaway: Compute $d$ carefully when terms are fractions.

PYQ 2: Which term is zero (2 marks)

Which term of the AP $5, 9, 13, \dots$ is 0? If none, explain. [CBSE Board]

Solution:

1. $a = 5$, $d = 4$. Set $5 + (n-1)4 = 0 \Rightarrow 4(n-1) = -5 \Rightarrow n - 1 = -\dfrac{5}{4}$.
2. $n$ is negative/fractional.

Final Answer: No term equals 0 (the AP has no zero term).

Takeaway: A non-positive-integer $n$ means the value is not a term.

PYQ 3: Find $a$ and $d$ (3 marks)

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

Solution:

1. $a_{17} - a_{10} = 7 \Rightarrow (a + 16d) - (a + 9d) = 7 \Rightarrow 7d = 7$.
2. $d = 1$.

Final Answer: $d = 1$.

Takeaway: $a_{17} - a_{10} = 7d$ gives $d$ directly.

PYQ 4: Sum of n terms (3 marks)

Find the sum of the first 24 terms of the AP $5, 8, 11, 14, \dots$ [CBSE Board]

Solution:

1. $a = 5$, $d = 3$. $S_{24} = \dfrac{24}{2}[10 + 23(3)] = 12[10 + 69] = 12(79) = 948$.

Final Answer: 948.

Takeaway: Apply $S_n = \dfrac{n}{2}[2a + (n-1)d]$.

PYQ 5: Find $n$ from a sum (3 marks)

How many terms of the AP $9, 17, 25, \dots$ must be taken so that their sum is 636? [CBSE Board]

Solution:

1. $a = 9$, $d = 8$. $\dfrac{n}{2}[18 + (n-1)8] = 636 \Rightarrow 4n^2 + 5n - 636 = 0$.
2. $(4n + 53)(n - 12) = 0 \Rightarrow n = 12$.

Final Answer: 12 terms.

Takeaway: Reject the negative root.

PYQ 6: Determine the AP (3 marks)

The 3rd term of an AP is 5 and the 7th term is 9. Find the AP and its 20th term. [CBSE Board]

Solution:

1. $a + 2d = 5$, $a + 6d = 9$. Subtract: $4d = 4 \Rightarrow d = 1$; $a = 3$.
2. AP: $3, 4, 5, \dots$; $a_{20} = 3 + 19(1) = 22$.

Final Answer: AP $3, 4, 5, \dots$; 20th term 22.

Takeaway: Two terms ⇒ $a$ and $d$ ⇒ any term.

PYQ 7: Word problem — logs (4 marks)

200 logs are stacked so the bottom row has 20 logs, the next 19, and so on. In how many rows are 200 logs placed, and how many are in the top row? [CBSE Board]

Solution:

1. $a = 20$, $d = -1$, $S_n = 200$. $\dfrac{n}{2}[40 - (n-1)] = 200 \Rightarrow n^2 - 41n + 400 = 0 \Rightarrow (n-16)(n-25) = 0$.
2. $n = 16$ (reject 25, which gives negative logs). Top row $= a_{16} = 20 + 15(-1) = 5$.

Final Answer: 16 rows; top row has 5 logs.

Takeaway: Reject the inadmissible root, then find the required term.

PYQ 8: Sum of two-digit multiples (3 marks)

Find the sum of all two-digit numbers which are divisible by 3. [CBSE Board]

Solution:

1. AP $12, 15, \dots, 99$. $a = 12$, $d = 3$, $l = 99$, $n = 30$ (since $99 = 12 + (n-1)3$).
2. $S_{30} = \dfrac{30}{2}(12 + 99) = 15(111) = 1665$.

Final Answer: 1665.

Takeaway: Find $n$, then use $\dfrac{n}{2}(a + l)$.

PYQ 9: Given a term and a sum (3 marks)

In an AP, the first term is 8, the nth term is 33, and the sum of the first $n$ terms is 123. Find $n$ and $d$. [CBSE Board]

Solution:

1. $S_n = \dfrac{n}{2}(a + a_n) = \dfrac{n}{2}(8 + 33) = \dfrac{41n}{2} = 123 \Rightarrow n = 6$.
2. $a_6 = 8 + 5d = 33 \Rightarrow 5d = 25 \Rightarrow d = 5$.

Final Answer: $n = 6$, $d = 5$.

Takeaway: Use $S_n = \dfrac{n}{2}(a + a_n)$ to find $n$ first.

PYQ 10: Sum of first n odd numbers (2 marks)

Find the sum of the first 15 multiples of 8. [CBSE Board]

Solution:

1. AP $8, 16, 24, \dots$: $a = 8$, $d = 8$, $n = 15$.
2. $S_{15} = \dfrac{15}{2}[16 + 14(8)] = \dfrac{15}{2}(128) = 960$.

Final Answer: 960.

Takeaway: Multiples of 8 form an AP with $a = d = 8$.

PYQ 11: nth term from end (2 marks)

Find the 8th term from the end of the AP $7, 10, 13, \dots, 184$. [CBSE Board]

Solution:

1. $l = 184$, $d = 3$. 8th from end $= l - (8-1)d = 184 - 21 = 163$.

Final Answer: 163.

Takeaway: $l - (n-1)d$ counts from the end.

PYQ 12: Show $S_n$ relation (3 marks)

The sum of the first $n$ terms of an AP is $S_n = 4n - n^2$. Find the first term, the sum of the first two terms, and the 10th term. [CBSE Board]

Solution:

1. $S_1 = a_1 = 4 - 1 = 3$; $S_2 = 8 - 4 = 4$.
2. $a_{10} = S_{10} - S_9 = (40 - 100) - (36 - 81) = -60 - (-45) = -15$.

Final Answer: $a_1 = 3$, $S_2 = 4$, $a_{10} = -15$.

Takeaway: $a_n = S_n - S_{n-1}$; $S_1$ gives the first term.

Note: also $a_n = S_n - S_{n-1} = 5 - 2n$, giving $a_{10} = -15$. ✓

PYQ 13: Three terms in AP (3 marks)

The sum of three numbers in AP is 12 and the sum of their cubes is 288. Find the numbers. [CBSE Board]

Solution:

1. $a - d, a, a + d$: $3a = 12 \Rightarrow a = 4$.
2. Cubes: $(4-d)^3 + 64 + (4+d)^3 = 288$. Now $(4-d)^3 + (4+d)^3 = 2(4^3 + 3 \cdot 4 d^2) = 128 + 24d^2$.
3. $128 + 24d^2 + 64 = 288 \Rightarrow 24d^2 = 96 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2$.
4. Numbers: $2, 4, 6$.

Final Answer: 2, 4, 6.

Takeaway: $(a-d)^3 + (a+d)^3 = 2a^3 + 6ad^2$ simplifies the cube sum.

PYQ 14: Word problem — savings (3 marks)

A sum of ₹2800 is to be used to give 10 cash prizes to students. If each prize is ₹40 less than its preceding prize, find the value of each prize. (Find the first prize.) [CBSE Board]

Solution:

1. Prizes form an AP with $d = -40$, $n = 10$, $S_{10} = 2800$.
2. $\dfrac{10}{2}[2a + 9(-40)] = 2800 \Rightarrow 5[2a - 360] = 2800 \Rightarrow 2a - 360 = 560 \Rightarrow a = 460$.

Final Answer: First prize ₹460 (then 420, 380, …).

Takeaway: Use $S_n$ with the known total to find the first term.

PYQ 15: Find $a$ from two sums (3 marks)

If $S_n$ denotes the sum of $n$ terms of an AP whose common difference is $d$ and first term is $a$, find $S_{12}$ if $a = 2$ and $d = 8$. [CBSE Board]

Solution:

1. $S_{12} = \dfrac{12}{2}[2(2) + 11(8)] = 6[4 + 88] = 6(92) = 552$.

Final Answer: 552.

Takeaway: Direct application of the sum formula.

PYQ 16: Middle term sum (3 marks)

Find the sum of the first 51 terms of an AP whose 2nd and 3rd terms are 14 and 18 respectively. [CBSE Board]

Solution:

1. $d = 18 - 14 = 4$; $a_2 = a + d = 14 \Rightarrow a = 10$.
2. $S_{51} = \dfrac{51}{2}[20 + 50(4)] = \dfrac{51}{2}(220) = 51(110) = 5610$.

Final Answer: 5610.

Takeaway: Find $a$ and $d$ from consecutive terms, then the sum.

PYQ 17: Which term exceeds (3 marks)

Determine the AP whose 4th term is 18 and the difference of the 9th term from the 15th term is 30. [CBSE Board]

Solution:

1. $a_{15} - a_9 = 6d = 30 \Rightarrow d = 5$.
2. $a_4 = a + 3d = 18 \Rightarrow a + 15 = 18 \Rightarrow a = 3$.
3. AP: $3, 8, 13, 18, \dots$

Final Answer: $3, 8, 13, 18, \dots$

Takeaway: $a_{15} - a_9 = 6d$; then use a known term for $a$.

PYQ 18: Sum of part of AP (3 marks)

The sum of the first $n$ terms of an AP is $3n^2 + 5n$. Find the AP and its 25th term. [CBSE Board]

Solution:

1. $a_n = S_n - S_{n-1} = 6n + 2$ (as derived earlier). $a_1 = 8$, $d = 6$.
2. AP: $8, 14, 20, \dots$; $a_{25} = 6(25) + 2 = 152$.

Final Answer: AP $8, 14, 20, \dots$; 25th term 152.

Takeaway: $a_n = S_n - S_{n-1}$ gives the general term.

PYQ 19: Equal sums (4 marks)

If the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289, find the sum of its first $n$ terms. [CBSE Board]

Solution:

1. $S_7 = 7(a + 3d) = 49 \Rightarrow a + 3d = 7$.
2. $S_{17} = 17(a + 8d) = 289 \Rightarrow a + 8d = 17$.
3. Subtract: $5d = 10 \Rightarrow d = 2$, $a = 1$.
4. $S_n = \dfrac{n}{2}[2 + (n-1)2] = n^2$.

Final Answer: $S_n = n^2$.

Takeaway: Two sum-conditions give $a$ and $d$, then general $S_n$.

PYQ 20: Word problem — production (3 marks)

A manufacturer produced 600 sets in the 3rd year and 700 in the 7th year. Assuming uniform increase, find the production in the 10th year and the total in the first 10 years. [CBSE Board]

Solution:

1. $a + 2d = 600$, $a + 6d = 700 \Rightarrow d = 25$, $a = 550$.
2. $a_{10} = 550 + 9(25) = 775$.
3. $S_{10} = \dfrac{10}{2}[1100 + 9(25)] = 5(1325) = 6625$.

Final Answer: 775 sets in year 10; 6625 in total.

Takeaway: Find $a$, $d$ first; then both the term and the sum.

PYQ 21: First term and ratio (3 marks)

The 5th term of an AP is 26 and the 10th term is 51. Find the AP. [CBSE Board]

Solution:

1. $a + 4d = 26$, $a + 9d = 51$. Subtract: $5d = 25 \Rightarrow d = 5$; $a = 6$.
2. AP: $6, 11, 16, 21, \dots$

Final Answer: $6, 11, 16, 21, \dots$

Takeaway: Standard two-term method.

PYQ 22: Sum of all terms (3 marks)

Find the sum of all natural numbers between 100 and 200 which are divisible by 4. [CBSE Board]

Solution:

1. AP $104, 108, \dots, 196$. $a = 104$, $d = 4$, $l = 196$. $196 = 104 + (n-1)4 \Rightarrow n = 24$.
2. $S = \dfrac{24}{2}(104 + 196) = 12(300) = 3600$.

Final Answer: 3600.

Takeaway: Identify the first and last multiples within the range.

PYQ 23: Term ratio condition (3 marks)

The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term. [CBSE Board]

Solution:

1. $a_4 = a + 3d = 0 \Rightarrow a = -3d$.
2. $a_{25} = a + 24d = -3d + 24d = 21d$; $a_{11} = a + 10d = -3d + 10d = 7d$.
3. $a_{25} = 21d = 3(7d) = 3 a_{11}$.

Final Answer: $a_{25} = 3 a_{11}$. $\blacksquare$

Takeaway: Express everything via $d$ using the condition $a = -3d$.

PYQ 24: Sum to a target (savings) (4 marks)

A contractor pays a penalty for delay: ₹200 for the first day, ₹250 for the second, ₹300 for the third, and so on. How much penalty for a 30-day delay? [CBSE Board]

Solution:

1. $a = 200$, $d = 50$, $n = 30$.
2. $S_{30} = \dfrac{30}{2}[400 + 29(50)] = 15[400 + 1450] = 15(1850) = 27750$.

Final Answer: ₹27750.

Takeaway: Daily penalties in AP; total is $S_n$.

PYQ 25: Find n given last term and sum (3 marks)

The first and 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? [CBSE Board]

Solution:

1. $350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow n = 38$.
2. $S_{38} = \dfrac{38}{2}(17 + 350) = 19(367) = 6973$.

Final Answer: 38 terms; sum 6973.

Takeaway: Use $l = a + (n-1)d$ for $n$, then $\dfrac{n}{2}(a + l)$.

PYQ 26: Combined conditions (4 marks)

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

Solution:

1. $2a + 10d = 24$ and $2a + 14d = 44$. Subtract: $4d = 20 \Rightarrow d = 5$; $a = -13$.
2. $S_{10} = \dfrac{10}{2}[2(-13) + 9(5)] = 5[-26 + 45] = 5(19) = 95$.

Final Answer: 95.

Takeaway: Form two equations from term-sums, then compute the required sum.