Given: AP has 40 terms, $S_9 = 153$, sum of last 6 terms = 687.

Step 1: Using $S_9 = 153$

$$S_9 = \frac{9}{2}[2a + 8d] = 153$$
$$2a + 8d = 34 \implies a + 4d = 17 \quad \text{...(i)}$$

Step 2: Sum of last 6 terms = 687

Sum of last 6 terms = $S_{40} - S_{34} = 687$

$$S_{40} = \frac{40}{2}[2a + 39d] = 20(2a + 39d)$$
$$S_{34} = \frac{34}{2}[2a + 33d] = 17(2a + 33d)$$

$$20(2a + 39d) - 17(2a + 33d) = 687$$
$$40a + 780d - 34a - 561d = 687$$
$$6a + 219d = 687 \implies 2a + 73d = 229 \quad \text{...(ii)}$$

Step 3: Solving (i) and (ii)

From (i): $a = 17 - 4d$. Substituting in (ii):
$$2(17 - 4d) + 73d = 229$$
$$34 + 65d = 229$$
$$d = 3$$
$$a = 17 - 12 = 5$$

First term $a = 5$, common difference $d = 3$

Step 4: Sum of all 40 terms

$$S_{40} = \frac{40}{2}[2(5) + 39(3)] = 20[10 + 117] = 20 \times 127 = \mathbf{2540}$$

Source: Chapter 5, Section 5.3 (Sum of first n terms of an AP)

---

• Key formula used: $S_n = \frac{n}{2}[2a + (n-1)d]$
• Examiner tip: "Sum of last 6 terms" must be expressed as $S_{40} - S_{34}$ — this is the most common error students make.
• Solve the two linear equations carefully; substitution is the safest method here.
• All three results (a, d, and $S_{40}$) must be stated clearly to score full marks. Each step carries marks, so show all working.

Let the first term = $a$ and common difference = $d$.

8th term: $a_8 = a + 7d$

Given:
$$a + a_8 = 32 \Rightarrow a + (a + 7d) = 32 \Rightarrow 2a + 7d = 32 \quad ...(1)$$
$$a \cdot a_8 = 60 \Rightarrow a(a + 7d) = 60 \quad ...(2)$$

From (1): $a + (a + 7d) = 32$, so let $a = x$ and $a + 7d = y$, where $x + y = 32$ and $xy = 60$.

These are roots of: $t^2 - 32t + 60 = 0$
$$t = \frac{32 \pm \sqrt{1024 - 240}}{2} = \frac{32 \pm \sqrt{784}}{2} = \frac{32 \pm 28}{2}$$

So $t = 30$ or $t = 2$.

Case 1: $a = 30$, $a + 7d = 2 \Rightarrow d = -4$

Case 2: $a = 2$, $a + 7d = 30 \Rightarrow d = 4$

Sum of first 20 terms:
$$S_{20} = \frac{20}{2}[2a + 19d]$$

• Case 1: $S_{20} = 10[60 + 19(-4)] = 10[60 - 76] = 10 \times (-16) = \mathbf{-160}$
• Case 2: $S_{20} = 10[4 + 19(4)] = 10[4 + 76] = 10 \times 80 = \mathbf{800}$

Source: Arithmetic Progressions, Chapter 5

---

• The key step is treating $a$ and $a_8$ as two unknowns with known sum (32) and product (60), forming a quadratic equation.
• Both cases are valid — examiners expect both solutions with corresponding $S_{20}$.
• Use $S_n = \frac{n}{2}[2a + (n-1)d]$ correctly for each case.
• Don't forget to verify: $30 \times 2 = 60$ ✓ and $30 + 2 = 32$ ✓.

Using $S_n = \dfrac{n}{2}(a + a_n)$:

$$\frac{2093}{2} = \frac{n}{2}(7 + 84) = \frac{91n}{2} \implies n = \frac{2093}{91} = 23$$

(c) 23

Use the shortcut formula $S_n = \frac{n}{2}(a + l)$ when first term, last term, and sum are all given. Dividing 2093 by 91 directly gives n = 23. No need to find d separately.

Given: $S_n = 2n^2 + 13n$

Finding the nth term:

$$a_n = S_n - S_{n-1}$$

$$S_{n-1} = 2(n-1)^2 + 13(n-1) = 2n^2 - 4n + 2 + 13n - 13 = 2n^2 + 9n - 11$$

$$a_n = (2n^2 + 13n) - (2n^2 + 9n - 11)$$

$$\boxed{a_n = 4n + 11}$$

Finding the 10th term:

$$a_{10} = 4(10) + 11 = 40 + 11 = \boxed{51}$$

Source: Chapter 5, Exercise 5.3

---

• The key formula is $a_n = S_n - S_{n-1}$, valid for $n \geq 2$. Examiners expect you to show this substitution step clearly.
• Always simplify $S_{n-1}$ carefully by expanding $(n-1)^2$ and $(n-1)$.
• Finally substitute $n = 10$ in $a_n$ to get the 10th term. Write both results clearly boxed or underlined for the examiner.

Step 1: Use the condition that 15th term exceeds 8th term by 21.

$a_{15} - a_8 = 21$

$[a + 14d] - [a + 7d] = 21$

$7d = 21 \Rightarrow d = 3$

Step 2: Use $S_{10} = 55$.

$S_{10} = \dfrac{10}{2}[2a + 9d] = 55$

$5[2a + 27] = 55$

$2a + 27 = 11$

$2a = -16 \Rightarrow a = -8$

Step 3: Form the A.P.

$a = -8,\ d = 3$

A.P.: –8, –5, –2, 1, 4, 7, ...

Source: Chapter 5 (Arithmetic Progressions), Exercise 5.2 & 5.3

---

• Examiners award 1 mark for finding $d$, 1 mark for finding $a$, and 1 mark for writing the A.P.
• Use $a_n = a + (n-1)d$ and $S_n = \frac{n}{2}[2a + (n-1)d]$ — both standard formulas from the chapter.
• Always write at least 4–5 terms of the A.P. at the end to confirm the answer.

Here, $d = -\frac{10}{4} - (-\frac{15}{4}) = \frac{5}{4}$. So $a_{16} - a_{12} = (16-12) \times d = 4 \times \frac{5}{4} = \mathbf{5}$.

(C) 5

Since $a_n - a_m = (n-m)d$ for any AP, you only need the common difference and the difference in term numbers. No need to calculate individual terms separately. Common difference here is $\frac{5}{4}$, and $16 - 12 = 4$, giving $4 \times \frac{5}{4} = 5$.

The 8th term from the end (of 30 terms) = the (30 − 8 + 1) = 23rd term from the beginning.

Given: $a_{30} = 119$ and $a_{23} = 91$

Using $a_n = a + (n-1)d$:

$$a_{30} - a_{23} = 7d \implies 119 - 91 = 7d \implies 28 = 7d \implies \boxed{d = 4}$$

Finding first term $a$:
$$119 = a + 29 \times 4 \implies a = 119 - 116 = 3$$

Sum of all 30 terms:
$$S_{30} = \frac{n}{2}(a + l) = \frac{30}{2}(3 + 119) = 15 \times 122 = \boxed{1830}$$

Source: Chapter 5, Sections 5.2 and 5.3

---

• Key step: The 8th term from the end in a 30-term AP is the $(30 - 8 + 1) = 23$rd term from the start. Students often miss this conversion — it's where most marks are lost.
• Use the difference of two $a_n$ expressions to eliminate $a$ and solve for $d$ directly.
• For sum, use $S = \frac{n}{2}(a + l)$ since both first and last terms are known — it's quicker than the other formula.
• Show each step clearly; all 3 marks are awarded for: finding the 23rd term position (½), finding $d$ (1), finding $S_{30}$ (1).

Numbers divisible by 5 but not by 2 are odd multiples of 5: 5, 15, 25, 35, …, 995.

This is an AP with $a = 5$, $d = 10$, $a_n = 995$.

$$a_n = a + (n-1)d \Rightarrow 995 = 5 + (n-1) \times 10$$

$$990 = (n-1) \times 10 \Rightarrow n - 1 = 99 \Rightarrow n = 100$$

Therefore, there are 100 such natural numbers.

Source: Chapter 5, Section 5.2 (nth term of an AP)

---

• Numbers divisible by 5 but not by 2 must be odd multiples of 5 (i.e., 5, 15, 25, … — skip even multiples like 10, 20, 30…).
• This forms an AP with common difference 10 (every alternate multiple of 5).
• Apply the nth term formula $a_n = a + (n-1)d$ to find $n$.
• Examiner expects you to identify the AP correctly and show the formula step clearly.

Given: $a_n = 5 - 2n$

First term: $a_1 = 5 - 2(1) = 3$

Common difference: $d = a_2 - a_1 = [5-2(2)] - 3 = 1 - 3 = -2$

Sum of first 20 terms:

$$S_{20} = \frac{n}{2}[2a + (n-1)d] = \frac{20}{2}[2(3) + (20-1)(-2)]$$

$$= 10[6 + 19 \times (-2)] = 10[6 - 38] = 10 \times (-32) = \mathbf{-320}$$

Source: Chapter 5, Exercise 5.3

---

• First find $a$ and $d$ from the given $a_n$ formula before applying the sum formula.
• The examiner expects you to show the values of $a$, $d$, and then substitute into $S_n = \frac{n}{2}[2a+(n-1)d]$.
• A common mistake is forgetting the negative sign in $d = -2$, which changes the answer entirely.

For $a, 7, b, 23$ to be in A.P., the common difference $d$ must be constant.

Common difference: $d = \dfrac{23 - 7}{3} = \dfrac{16}{3}$

(Since there are 3 equal steps from 7 to 23 via $b$: positions 2, 3, 4)

$$a = 7 - d = 7 - \frac{16}{3} = \frac{21 - 16}{3} = \frac{5}{3}$$

$$b = 7 + d = 7 + \frac{16}{3} = \frac{21 + 16}{3} = \frac{37}{3}$$

Therefore, $a = \dfrac{5}{3}$ and $b = \dfrac{37}{3}$.

Source: Chapter 5, Exercise 5.2

---

• The four terms occupy positions 1, 2, 3, 4, so from term 2 (which is 7) to term 4 (which is 23), there are exactly 2 steps of $d$, giving $2d = 16$, so $d = 8$.

Wait — recheck: positions are 1→2→3→4, so $23 = 7 + 2d \Rightarrow d = 8$.

Correction to model answer:

$d = \frac{23-7}{2} = 8$; $a = 7 - 8 = -1$; $b = 7 + 8 = 15$.

The examiner expects: $a = -1$, $b = 15$. Use the fact that 7 is the 2nd term and 23 is the 4th term, so $23 - 7 = 2d$.

Given AP: $-\dfrac{11}{2}, -3, -\dfrac{1}{2}, \ldots$

Here, $a = -\dfrac{11}{2}$, $d = -3 - \left(-\dfrac{11}{2}\right) = -3 + \dfrac{11}{2} = \dfrac{5}{2}$

Let $a_n = \dfrac{49}{2}$

Using $a_n = a + (n-1)d$:

$$\frac{49}{2} = -\frac{11}{2} + (n-1)\cdot\frac{5}{2}$$

$$\frac{49}{2} + \frac{11}{2} = (n-1)\cdot\frac{5}{2}$$

$$30 = (n-1)\cdot\frac{5}{2} \implies n-1 = 12 \implies n = 13$$

$\therefore \dfrac{49}{2}$ is the 13th term of the AP.

Source: Chapter 5, Exercise 5.2

---

• Identify $a$ and $d$ first; here both involve fractions, so be careful with subtraction.
• Set $a_n$ equal to the given value and solve for $n$.
• Examiners award 1 mark for correct $a$ and $d$, and 1 mark for the correct value of $n$. Show all steps clearly.

Given: $S_6 = 36$, $S_{16} = 256$

Using formula $S_n = \dfrac{n}{2}[2a + (n-1)d]$

From $S_6 = 36$:
$$\frac{6}{2}[2a + 5d] = 36$$
$$3[2a + 5d] = 36$$
$$2a + 5d = 12 \quad \text{...(i)}$$

From $S_{16} = 256$:
$$\frac{16}{2}[2a + 15d] = 256$$
$$8[2a + 15d] = 256$$
$$2a + 15d = 32 \quad \text{...(ii)}$$

Subtracting (i) from (ii):
$$10d = 20 \implies d = 2$$

Substituting in (i):
$$2a + 5(2) = 12 \implies 2a = 2 \implies a = 1$$

Sum of first 10 terms:
$$S_{10} = \frac{10}{2}[2(1) + 9(2)] = 5[2 + 18] = 5 \times 20 = \boxed{100}$$

Source: Chapter 5, Exercise 5.3

---

• Examiners award marks at each step: setting up both equations (1 mark each), solving for $d$ and $a$ (1 mark each), and finding $S_{10}$ (1 mark) — total 5 marks.
• Always write the formula first, substitute clearly, and show subtraction of equations explicitly — these are scoring steps.
• This question is structurally identical to Q.9 of Exercise 5.3 (sum of $n$ terms); practise that pattern.

Let the first term be $a$ and common difference be $d$.

Given: $\dfrac{a_{11}}{a_{18}} = \dfrac{2}{3}$

$$\frac{a + 10d}{a + 17d} = \frac{2}{3}$$

$$3(a + 10d) = 2(a + 17d)$$

$$3a + 30d = 2a + 34d$$

$$a = 4d$$

Ratio of 5th term to 21st term:

$$\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3}$$

$$\boxed{a_5 : a_{21} = 1 : 3}$$

Ratio of sum of first 5 terms to sum of first 21 terms:

$$S_5 = \frac{5}{2}[2a + 4d] = \frac{5}{2}[8d + 4d] = \frac{5}{2} \times 12d = 30d$$

$$S_{21} = \frac{21}{2}[2a + 20d] = \frac{21}{2}[8d + 20d] = \frac{21}{2} \times 28d = 294d$$

$$\frac{S_5}{S_{21}} = \frac{30d}{294d} = \frac{5}{49}$$

$$\boxed{S_5 : S_{21} = 5 : 49}$$

Source: Chapter 5, Section 5.3 (nth term and Sum of AP)

---

• Examiners award marks for: setting up the ratio equation correctly (1 mark), solving to get $a = 4d$ (1 mark), finding $a_5 : a_{21}$ (1 mark), and finding $S_5 : S_{21}$ (2 marks).
• Key formulas: $a_n = a + (n-1)d$ and $S_n = \frac{n}{2}[2a + (n-1)d]$.
• Substitute $a = 4d$ throughout — this is the critical step. Show substitution clearly for full marks.

$S_1 = 3(1)^2 + 1 = 4$. Since $S_1 = a_1$, the first term is 4.

(D) 4

The first term of any AP equals $S_1$ (sum of first 1 term). Substitute $n = 1$ in $S_n = 3n^2 + n$: $S_1 = 3 + 1 = 4$. You can verify: $a_2 = S_2 - S_1 = (12+2)-(3+1) = 10$, and $a_2 - a_1 = 10 - 4 = 6$ ✓ (matches given $d = 6$).

The number of logs in each row forms an AP:

• First term $a = 22$, common difference $d = -1$, Total $S_n = 250$

Finding number of rows (n):

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$

$$250 = \frac{n}{2}[2(22) + (n-1)(-1)]$$

$$500 = n[44 - n + 1] = n[45 - n]$$

$$500 = 45n - n^2$$

$$n^2 - 45n + 500 = 0$$

$$(n - 20)(n - 25) = 0$$

So $n = 20$ or $n = 25$.

Finding the top row for each value of n:

If $n = 25$: $a_{25} = 22 + (25-1)(-1) = 22 - 24 = -2$

Since number of logs cannot be negative, $n = 25$ is rejected.

If $n = 20$: $a_{20} = 22 + (20-1)(-1) = 22 - 19 = 3$

∴ The 250 logs are placed in 20 rows and the top row has 3 logs.

Source: Chapter 5, Section 5.4

---

• Identify this as an AP sum problem: rows decrease by 1 log each time, so $d = -1$.
• Set up $S_n = 250$ and solve the quadratic — you get two values of $n$.
• Always check both values: $n = 25$ gives a negative number of logs in the top row, which is impossible, so reject it.
• Examiners award marks for: correct AP identification, correct formula use, solving the quadratic, rejecting $n = 25$ with reason, and stating the final answer clearly.

Let the first term be $a$ and common difference be $d$.

Step 1: Use given ratio of 11th and 17th terms

$$\frac{a_{11}}{a_{17}} = \frac{3}{4}$$

$$\frac{a + 10d}{a + 16d} = \frac{3}{4}$$

$$4(a + 10d) = 3(a + 16d)$$

$$4a + 40d = 3a + 48d$$

$$a = 8d$$

Step 2: Ratio of 5th term to 21st term

$$a_5 = a + 4d = 8d + 4d = 12d$$

$$a_{21} = a + 20d = 8d + 20d = 28d$$

$$\frac{a_5}{a_{21}} = \frac{12d}{28d} = \boxed{\frac{3}{7}}$$

Step 3: Ratio of sum of first 5 terms to sum of first 21 terms

$$S_5 = \frac{5}{2}(a + a_5) = \frac{5}{2}(8d + 12d) = \frac{5}{2} \times 20d = 50d$$

$$S_{21} = \frac{21}{2}(a + a_{21}) = \frac{21}{2}(8d + 28d) = \frac{21}{2} \times 36d = 378d$$

$$\frac{S_5}{S_{21}} = \frac{50d}{378d} = \boxed{\frac{25}{189}}$$

Source: Chapter 5 (Arithmetic Progressions), Section 5.3–5.4

---

• Examiners award marks at each step: setting up the equation (1M), solving for $a$ in terms of $d$ (1M), finding term ratio (1M), setting up sum formula (1M), and final sum ratio (1M).
• Use $S_n = \frac{n}{2}(a + l)$ where $l$ is the last term — it's faster than the $2a+(n-1)d$ form here.
• Always simplify the final ratio fully; leaving it unsimplified loses the accuracy mark.
• The key insight is expressing $a = 8d$ so every term cancels to a pure number ratio.

Option (C) 0

For an AP, the common difference is constant: $(p+1)-(p-1) = (2p+3)-(p+1)$
$\Rightarrow 2 = p+2 \Rightarrow p = 0$

Use the AP condition: middle term – first term = third term – middle term, i.e., $2b = a + c$. Set the two consecutive differences equal and solve. A common error is arithmetic mistakes while simplifying — write each step clearly.

Given: a = 400 m, d = 7.6 m (AP of lane lengths)

(i) Length of 6th lane:

$a_6 = a + 5d = 400 + 5(7.6) = 400 + 38 = \mathbf{438\ m}$

(ii) How much longer is 8th lane than 4th lane:

$a_8 = 400 + 7(7.6) = 400 + 53.2 = 453.2\ m$

$a_4 = 400 + 3(7.6) = 400 + 22.8 = 422.8\ m$

Difference $= 453.2 - 422.8 = \mathbf{30.4\ m}$

(iii) Total distance covered in first six lanes:

$S_6 = \frac{6}{2}[2a + 5d] = 3[2(400) + 5(7.6)] = 3[800 + 38] = 3 \times 838 = \mathbf{2514\ m}$

Source: Arithmetic Progressions, Application-based case study

---

• Lane lengths form an AP with first term a = 400, common difference d = 7.6.
• For (ii), the difference between any two lanes = (difference in their positions) × d = 4 × 7.6 = 30.4 m — you can use this shortcut.
• For (iii), use sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with n = 6. Examiners award 1 mark for correct formula and 1 mark for correct calculation.

The number of triangles in each row forms an AP: 1, 3, 5, 7, … with first term $a = 1$ and common difference $d = 2$.

(i) The bottom-most row is the 10th row.

$a_{10} = 1 + (10-1) \times 2 = 1 + 18 = \mathbf{19}$

There are 19 triangles in the bottom-most row.

(ii) The 4th row from the bottom is the 7th row (from top).

$a_7 = 1 + (7-1) \times 2 = 1 + 12 = \mathbf{13}$

There are 13 triangles in the 4th row from the bottom.

(iii) Total triangles up to 8th row:

$$S_8 = \frac{8}{2}[2(1) + (8-1) \times 2] = 4[2 + 14] = 4 \times 16 = \mathbf{64}$$

Total number of triangles till 8th row = 64.

---

• The rows from top form an AP with $a=1$, $d=2$, so the $n$th row has $(2n-1)$ triangles.
• For (ii), "4th from bottom" in a 10-row triangle = row number $10 - 4 + 1 = 7$ from top.
• For (iii), use the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$. Examiners expect all steps shown clearly.

Option (A)

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Verification: For AP 3, 6, 9, …, 198: $l = 198$, $d = 3$, so 10th term from end $= 198 - (10-1) \times 3 = 198 - 27 = 171$.

Wait — the calculated value is 171, not 168, so Assertion (A) is false.

Correct Option: D — Assertion (A) is false, but Reason (R) is true.

• The formula in R is correct: nth term from end $= l - (n-1)d$.
• Applying it: $198 - 9 \times 3 = 198 - 27 = 171 \neq 168$, so A is false.
• Always verify the assertion numerically before choosing option A; do not assume both are true just because the formula looks right.

Given: Total length = 5000 m, first pole at 200 m from base, common difference d = 150 m, last pole is 300 m from top.

The AP for pole distances from base: a = 200, d = 150.

(i) Distance of 10th pole from base:

$a_{10} = a + 9d = 200 + 9(150) = 200 + 1350 = \textbf{1550 m}$

(ii) Distance between 15th and 25th pole:

$a_{15} = 200 + 14(150) = 2300$ m

$a_{25} = 200 + 24(150) = 3800$ m

Distance = $3800 - 2300 = \textbf{1500 m}$

(iii) Time taken to reach 15th pole from the top:

Distance of 15th pole from base = 2300 m

Distance of 15th pole from top = $5000 - 2300 = 2700$ m

Time = $\dfrac{2700}{5} = \textbf{540 seconds}$

Source: Arithmetic Progressions, Case Study

---

• The AP is straightforward: a = 200, d = 150. Use $a_n = a + (n-1)d$.
• For (ii), the difference between any two consecutive 10 poles is simply $10 \times 150 = 1500$ m — a quicker check.
• For (iii), examiners expect you to find the distance from the top, not from the base. Subtract the pole's distance from the total 5000 m, then divide by speed.
• The information about "last pole 300 m from top" can be used to verify the number of poles but isn't needed for these sub-questions.

Given: $a_n = 3n - 2$

First term: $a_1 = 3(1) - 2 = 1$

Last term (28th): $a_{28} = 3(28) - 2 = 84 - 2 = 82$

Using $S_n = \dfrac{n}{2}(a + l)$:

$$S_{28} = \frac{28}{2}(1 + 82) = 14 \times 83 = \mathbf{1162}$$

Source: Chapter 5, Summary Point 5

• Find first term by putting $n = 1$, and last (28th) term by putting $n = 28$.
• Use the formula $S = \frac{n}{2}(a + l)$ since both first and last terms are known — this is quicker than using $S = \frac{n}{2}[2a + (n-1)d]$.
• Both approaches are acceptable; examiners award marks for correct formula, substitution, and final answer.

Given: $a = 32$, $l = -10$, $d = -2$

Finding n:
Using $a_n = a + (n-1)d$:
$$-10 = 32 + (n-1)(-2)$$
$$-42 = -2(n-1)$$
$$n - 1 = 21 \Rightarrow n = 22$$

Finding Sum:
$$S = \frac{n}{2}(a + l) = \frac{22}{2}(32 + (-10)) = 11 \times 22 = 242$$

∴ Number of terms = 22 and Sum = 242

Source: Chapter 5, Summary (points 3 and 5)

---

• Use $a_n = a + (n-1)d$ to find $n$ first (since last term $= a_n$).
• Then use $S = \frac{n}{2}(a + l)$ — the shorter sum formula when both first and last terms are known. This saves calculation time.
• Both steps are needed for full 2 marks; losing either costs 1 mark.

Here, $a = -\dfrac{1}{3}$, $d = \dfrac{2}{3} - \left(-\dfrac{1}{3}\right) = 1$.

$a_n = a + (n-1)d = -\dfrac{1}{3} + (n-1)(1) = n - 1 - \dfrac{1}{3} = n - \dfrac{4}{3}$

Answer: (B) $n - \dfrac{4}{3}$

Use $a_n = a + (n-1)d$ with $a = -\frac{1}{3}$ and $d = 1$. Simplify carefully: $-\frac{1}{3} + n - 1 = n - \frac{4}{3}$. The trap options mix up signs or place the fraction incorrectly — always simplify step by step. Source: Chapter 5, Section 5.3 (nth term formula).

The multiples of 4 between 12 and 250 form the AP: 16, 20, 24, …, 248 with $a = 16$, $d = 4$, $a_n = 248$.

$248 = 16 + (n-1) \times 4 \Rightarrow 232 = (n-1) \times 4 \Rightarrow n = 59$

Answer: (A) 59

Source: Exercise 5.2, Q.14, Chapter 5

The key is identifying the first multiple of 4 strictly between 12 and 250 (which is 16, not 12) and the last (which is 248, not 252). Applying $a_n = a + (n-1)d$ gives $n = 59$. Students often err by including 12 or 252, getting 60 or 61. The textbook Exercise 5.2 Q.14 asks a similar question ("between 10 and 250"), giving 59 — confirming option A here.

The instalments form an AP where a = ₹1000, d = ₹100.

(i) 30th instalment:

$a_{30} = a + 29d = 1000 + 29 \times 100 = 1000 + 2900 = ₹3900$

(ii) Last (40th) instalment:

$a_{40} = 1000 + 39 \times 100 = 1000 + 3900 = ₹4900$

(iii) Amount paid after 30 instalments:

$S_{30} = \frac{30}{2}[2 \times 1000 + 29 \times 100] = 15 \times [2000 + 2900] = 15 \times 4900 = ₹73,500$

Amount still to be paid $= 1,18,000 - 73,500 = ₹44,500$

Source: Arithmetic Progressions, Application of AP (Sum of n terms)

---

• Key formula used: $a_n = a + (n-1)d$ for the nth term, and $S_n = \frac{n}{2}[2a + (n-1)d]$ for sum.
• For part (iii), examiners expect you to find $S_{30}$ first, then subtract from the total loan — both steps carry marks.
• Write the AP identification (a, d) clearly at the start to show your method.

(D) Assertion (A) is false, but Reason (R) is true.

The sum of first n natural numbers is $\frac{n(n+1)}{2}$ (R is true), so Mean $= \frac{n(n+1)/2}{n} = \frac{n+1}{2}$, not $\frac{n-1}{2}$. Hence A is false.

• R is true: The standard formula $1+2+\cdots+n = \frac{n(n+1)}{2}$ is correct.
• A is false: Dividing the sum by n gives mean $= \frac{n+1}{2}$, not $\frac{n-1}{2}$ (e.g., mean of 1, 2, 3 is 2 = $\frac{3+1}{2}$, not $\frac{3-1}{2} = 1$).
• In Assertion–Reason questions, always verify A independently before checking if R explains it.

Option D: $1^2, 3^2, 5^2, 7^2, \ldots$ i.e., $1, 9, 25, 49, \ldots$ is not an AP, as the differences $9-1=8$, $25-9=16$, $49-25=24$ are not equal.

Check each option by finding consecutive differences:

• A: $d = \frac{1}{2}$ ✓ (AP)
• B: $d = -2$ ✓ (AP)
• C: $\sqrt{2}, \sqrt{8}=2\sqrt{2}, \sqrt{18}=3\sqrt{2}$ → $d = \sqrt{2}$ ✓ (AP)
• D: $1, 9, 25, 49$ → differences 8, 16, 24 — not constant ✗ (NOT an AP)

The textbook (Exercise 5.1, Q4, xiv) confirms $1^2, 3^2, 5^2, 7^2$ is an AP — but that question asks to find common difference if it is one. Actually, verifying shows it is not an AP since differences are unequal. Option D is the correct answer.

(i) The first potato is 5 m from the bucket. The competitor runs 5 m to pick it up and 5 m to return.
Distance = 5 + 5 = 10 m

(ii) The second potato is 5 + 3 = 8 m from the bucket. The competitor runs 8 m and returns 8 m.
Distance = 8 + 8 = 16 m

(iii) Distance covered for the $n$-th potato = $2[5 + (n-1) \times 3]$ m

So the distances form an AP: 10, 16, 22, … with $a = 10$, $d = 6$, $n = 10$.

Total distance $S_{10} = \dfrac{10}{2}[2(10) + (10-1)(6)]$

$= 5[20 + 54] = 5 \times 74 = \mathbf{370 \text{ m}}$

Source: Arithmetic Progressions, Application/Case Study

---

• Each trip (to pick up potato $n$ and return) covers $2 \times$ (distance of that potato from bucket).
• Distances of potatoes from bucket: 5, 8, 11, … (AP with $a=5$, $d=3$), so trips form AP 10, 16, 22, … ($a=10$, $d=6$).
• Use $S_n = \frac{n}{2}[2a + (n-1)d]$ for the total. Examiners award 1 mark each for parts (i) and (ii), and for part (iii) they look for correct AP identification + correct formula application + final answer (2 marks).

(a) $\sqrt{2}$

Common difference $d = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$.

The common difference is found by subtracting any term from the next term. Here, each term increases by $\sqrt{2}$, so $d = \sqrt{2}$. Option (a) is correct.

For AP: 9, 7, 5, … → $a = 9$, $d = -2$

$a_n = 9 + (n-1)(-2) = 11 - 2n$

For AP: 15, 12, 9, … → $a = 15$, $d = -3$

$a_n = 15 + (n-1)(-3) = 18 - 3n$

Setting them equal:

$$11 - 2n = 18 - 3n$$

$$n = 7$$

∴ The 7th terms of both APs are equal.

Source: Chapter 5, Exercise 5.2 (Q.15 type); Section 5.3 – nth term formula

---

• Write the $n$th term formula $a_n = a + (n-1)d$ for each AP separately.
• Equate the two expressions and solve for $n$.
• Examiners award 1 mark for correctly setting up both $a_n$ expressions and 1 mark for solving $n = 7$.
• Always verify: $a_7$ for AP1 = $11-14 = -3$; $a_7$ for AP2 = $18-21 = -3$. ✓

Given: $d = 5$, $a_{20} = 135$, $n = 20$

Using $a_n = a + (n-1)d$:

$135 = a + 19 \times 5$

$a = 135 - 95 = 40$

Now, $S_{20} = \dfrac{n}{2}(a + l) = \dfrac{20}{2}(40 + 135) = 10 \times 175 = \mathbf{1750}$

Source: Chapter 5, Exercise 5.3

---

• First find the first term $a$ using $a_n = a + (n-1)d$.
• Then use $S = \frac{n}{2}(a + l)$ where $l = a_{20} = 135$ (the last term here), which is simpler than using $S = \frac{n}{2}[2a + (n-1)d]$.
• Both formulas give the same answer; examiners award 1 mark for finding $a$ and 1 mark for the correct sum.

(a)
Number of triangles in each figure: 1, 2, 3, 4, …

This is an AP with first term a = 1 and common difference d = 1.

The *n*th term: $a_n = a + (n-1)d = 1 + (n-1)(1) = n$

So, $a_n = n$

---

(b)
For triangles in a row, matchsticks used = $2n + 1$ (each new triangle adds 2 matchsticks to the previous figure, starting with 3).

AP for matchsticks: 3, 5, 7, … with a = 3, d = 2.

Given $a_n = 61$:

$$a + (n-1)d = 61$$
$$3 + (n-1)(2) = 61$$
$$(n-1)(2) = 58$$
$$n - 1 = 29 \Rightarrow n = 30$$

∴ Figure 30 has 61 matchsticks.

Source: Arithmetic Progressions, Linear Relations through Matchstick Patterns

---

• Part (a) asks for the AP of number of triangles (1, 2, 3…), not matchsticks — keep them separate.
• Part (b) uses the matchstick count AP: Figure 1 uses 3, Figure 2 uses 5, Figure 3 uses 7 (each new triangle shares one side, so adds 2 sticks). Formula $a_n = 2n+1$ must be derived from this pattern.
• Always show the substitution step clearly — it earns method marks.

Given: $S_n = n(4n + 1) = 4n^2 + n$

First term: $a_1 = S_1 = 4(1)^2 + 1 = 5$

Second term: $S_2 = 4(4) + 2 = 18$, so $a_2 = S_2 - S_1 = 18 - 5 = 13$

Third term: $S_3 = 4(9) + 3 = 39$, so $a_3 = S_3 - S_2 = 39 - 18 = 21$

Common difference $d = 13 - 5 = 8$

∴ The AP is 5, 13, 21, 29, …

Source: Chapter 5, Arithmetic Progressions

---

• Examiners expect you to find $a_1 = S_1$, then use $a_n = S_n - S_{n-1}$ for $n \geq 2$ to get at least two more terms.
• Always verify $d$ is constant to confirm it's an AP.
• Writing the AP at the end is essential for full marks — don't leave it implied.

Given AP: 30, 24, 18, …

Here, $a = 30$, $d = 24 - 30 = -6$, $n = 30$

Using the formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$

$$S_{30} = \frac{30}{2}[2(30) + (30-1)(-6)]$$

$$= 15[60 + 29 \times (-6)]$$

$$= 15[60 - 174]$$

$$= 15 \times (-114) = \mathbf{-1710}$$

Source: Chapter 5, Exercise 5.3

---

• Examiners award 1 mark for correctly identifying $a$, $d$, $n$ and writing the formula, and 1 mark for the correct final answer.
• The key here is noting that $d$ is negative (−6), which makes the sum negative — a common error is using $d = +6$.
• Always substitute carefully and simplify step by step to avoid sign mistakes.

Given: $p$th term $= q$ and $q$th term $= p$

Using $a_n = a + (n-1)d$:

$$a_p = a + (p-1)d = q \quad \text{...(1)}$$

$$a_q = a + (q-1)d = p \quad \text{...(2)}$$

Subtracting (2) from (1):

$$(p-1)d - (q-1)d = q - p$$

$$d(p - q) = -(p - q)$$

$$d = -1$$

Substituting $d = -1$ in (1):

$$a + (p-1)(-1) = q$$

$$a = q + p - 1 = p + q - 1$$

nth term:

$$a_n = a + (n-1)d = (p + q - 1) + (n-1)(-1)$$

$$a_n = p + q - 1 - n + 1 = \boxed{p + q - n}$$

Hence proved.

Source: Chapter 5, Section 5.3 (nth term formula $a_n = a + (n-1)d$)

---

• Examiners award marks for: setting up two equations (1 mark), finding $d = -1$ by subtraction (1 mark), and correctly arriving at $a_n = p+q-n$ (1 mark).
• The key trick is subtracting the two equations to eliminate $a$ and find $d$ first, then back-substitute to get $a$.
• Write each step clearly; do not skip the substitution step as it carries a mark.

The AP is 10, 7, 4, …, –62 with $d = -3$, last term $l = -62$.

The 11th term from the end = $l + (11-1)(-d)^{-1}$... Using: $n$th term from end $= l + (n-1) \times (-d)$

$$= -62 + (11-1)(3) = -62 + 30 = \mathbf{-32}$$

Answer: (c) –32

To find the $n$th term from the end of an AP, treat the last term as the first term and reverse the common difference: $T_n(\text{from end}) = l + (n-1)(-d)$. Here $l = -62$, $n = 11$, $-d = 3$, giving $-62 + 30 = -32$. Alternatively, find total number of terms first (25 terms), then the 11th from end is the 15th from the start: $a_{15} = 10 + 14(-3) = -32$.

(b) 3

$a_n = 3n + 7$, so $d = a_{n+1} - a_n = [3(n+1)+7] - [3n+7] = 3$.

The common difference is found by subtracting the $k$th term from the $(k+1)$th term. Since $a_n = 3n+7$ is linear in $n$, the coefficient of $n$ (i.e., 3) directly gives the common difference. The constant 7 is part of the first term, not $d$.

Given AP: 65, 61, 57, 53, …

Here, $a = 65$, $d = 61 - 65 = -4$

Using the formula: $a_n = a + (n-1)d$

For the first negative term, we need $a_n < 0$:

$$65 + (n-1)(-4) < 0$$

$$65 - 4n + 4 < 0$$

$$69 - 4n < 0$$

$$4n > 69$$

$$n > 17.25$$

Since $n$ must be a whole number, the smallest value is $n = 18$.

The 18th term is the first negative term.

Verification: $a_{18} = 65 + 17 \times (-4) = 65 - 68 = -3$ ✓

Source: Chapter 5, Section 5.3 (nth term formula)

---

• Set up $a_n < 0$ using $a_n = a + (n-1)d$ and solve the inequality for $n$.
• Since $n$ must be a natural number, always round up to the next whole number (here 17.25 → 18).
• Verify by substituting $n = 18$ to confirm the term is negative and check $n = 17$ gives a positive value ($a_{17} = 65 - 64 = 1 > 0$).
• Examiners expect the inequality approach, not trial-and-error.

Given: $a = -14$, $a_5 = 2$, $l = a_n = 62$

Step 1: Find common difference $d$

Using $a_5 = a + 4d$:
$$2 = -14 + 4d \implies 4d = 16 \implies d = 4$$

Step 2: Find number of terms $n$

Using $a_n = a + (n-1)d$:
$$62 = -14 + (n-1) \times 4$$
$$76 = (n-1) \times 4$$
$$n - 1 = 19$$
$$n = 20$$

∴ There are 20 terms in the AP.

Source: Chapter 5, Section 5.2 (nth term formula)

---

• First find $d$ using the 5th term condition — this is the key intermediate step examiners look for (1 mark).
• Then apply $a_n = a + (n-1)d$ with the last term to find $n$ (1 mark each for correct substitution and answer).
• Always state the formula before substituting — it earns a method mark even if arithmetic slips.

(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

A is true: As stated in the textbook (Note to the Reader, Ch. 5), if $a$, $b$, $c$ are in AP then $b = \frac{a+c}{2}$, i.e., $2b = a + c$. R is also true: The sum of first $n$ odd natural numbers = $n^2$ is a standard result. However, R has no connection to the condition for three terms to be in AP, so it does not explain A. Hence option (b).

The terms are $\sqrt{6}$, $\sqrt{24}$, $\sqrt{54}$, i.e., $\sqrt{6}$, $2\sqrt{6}$, $3\sqrt{6}$, … with common difference $\sqrt{6}$. Next term = $4\sqrt{6} = \sqrt{96}$. Answer: (b) $\sqrt{96}$

Rewrite each term: $\sqrt{6}=1\sqrt{6}$, $\sqrt{24}=2\sqrt{6}$, $\sqrt{54}=3\sqrt{6}$. The AP has $d=\sqrt{6}$, so the 4th term is $4\sqrt{6}=\sqrt{96}$. Always simplify surd terms to spot the pattern.

Given: AP = 45, 39, 33, …
Here, $a = 45$, $d = 39 - 45 = -6$, $S_n = 180$

Using the sum formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$

$$180 = \frac{n}{2}[2(45) + (n-1)(-6)]$$

$$360 = n[90 - 6n + 6]$$

$$360 = n[96 - 6n]$$

$$360 = 96n - 6n^2$$

$$6n^2 - 96n + 360 = 0$$

$$n^2 - 16n + 60 = 0$$

$$(n - 6)(n - 10) = 0$$

$$\therefore n = 6 \quad \text{or} \quad n = 10$$

Explanation of double answer:

Both answers are valid. When $n = 6$, $S_6 = 180$.
When $n = 10$, $S_{10} = 180$ also.

This happens because the AP has a positive first term but a negative common difference ($d = -6$), so the terms decrease and eventually become negative. The terms from 7th to 10th are negative (e.g., $a_7 = 45 + 6(-6) = 9$; $a_8 = 3$; $a_9 = -3$; $a_{10} = -9$). The negative terms from 7th onward cancel the extra sum added after the 6th term, making $S_{10} = S_6 = 180$.

Source: Chapter 5, Section 5.3 — Sum of First n Terms of an AP

---

• Key formula: $S_n = \frac{n}{2}[2a + (n-1)d]$; substituting correctly gives a quadratic in $n$.
• Both roots are positive integers, so both are accepted — write both.
• Examiner expects the double-answer explanation: The terms after $n = 6$ are small/negative (AP is decreasing), so their net contribution is zero over terms 7–10, giving the same sum again. Always verify both values of $n$ are natural numbers.
• Award of marks: ~2 marks for setting up and solving the quadratic, 1 mark each for the two values of $n$, 1 mark for the explanation.

Given: $a = 5$, $l = 45$, $S = 400$

Finding n:

Using $S = \dfrac{n}{2}(a + l)$

$$400 = \frac{n}{2}(5 + 45)$$

$$400 = \frac{n}{2} \times 50 = 25n$$

$$n = 16$$

Finding common difference d:

Using $a_n = a + (n-1)d$

$$45 = 5 + (16-1)d$$

$$40 = 15d$$

$$d = \frac{40}{15} = \frac{8}{3}$$

∴ Number of terms = 16 and common difference $d = \dfrac{8}{3}$

Source: Arithmetic Progressions, Exercise 5.3 Q5

---

• Examiners expect two clear steps: first find $n$ using the sum formula with last term $S = \frac{n}{2}(a+l)$, then find $d$ using the $n$th term formula $a_n = a + (n-1)d$.
• Write the formula, substitute values, and solve — show all working for full 3 marks.
• A common mistake is using $S = \frac{n}{2}[2a+(n-1)d]$ first (which has two unknowns); always use $S = \frac{n}{2}(a+l)$ when the last term is given.

Given: $S_{14} = 1050$, $a = 10$

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

$$1050 = \frac{14}{2}[2(10) + 13d]$$

$$1050 = 7[20 + 13d]$$

$$150 = 20 + 13d$$

$$d = \frac{130}{13} = 10$$

20th term:

$$a_{20} = a + 19d = 10 + 19(10) = 10 + 190 = \mathbf{200}$$

nth term:

$$a_n = a + (n-1)d = 10 + (n-1)(10) = 10 + 10n - 10 = \mathbf{10n}$$

Source: Chapter 5, Exercise 5.3 / Section 5.5

---

• First find $d$ using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$. Substituting $n = 14$, $S_{14} = 1050$, $a = 10$ gives $d = 10$.
• Then apply $a_n = a + (n-1)d$ for both $n = 20$ and general $n$.
• Examiners award marks for: correct formula used (1 mark), finding $d$ (1 mark), finding $a_{20}$ and $a_n$ (1 mark). Show all substitution steps clearly.

(C) 37

Here, $a = 3$, $d = 3$, $a_n = 111$. Using $a_n = a + (n-1)d$: $111 = 3 + (n-1)3 \Rightarrow 108 = 3(n-1) \Rightarrow n = 37$.

Use the formula $a_n = a + (n-1)d$ with the last term as $a_n$. Solve for $n$. A common mistake is forgetting to add 1 after dividing — always verify by checking $a_{37} = 3 + 36 \times 3 = 111$ ✓.

The terms simplify: $\sqrt{18}=3\sqrt{2}$, $\sqrt{50}=5\sqrt{2}$, $\sqrt{98}=7\sqrt{2}$, so $d=2\sqrt{2}$. The 4th term $=7\sqrt{2}+2\sqrt{2}=9\sqrt{2}=\sqrt{162}$. Answer: (C) $\sqrt{162}$

Rewrite each term as a multiple of $\sqrt{2}$: the AP is $3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}, \ldots$ with common difference $2\sqrt{2}$. The 4th term is $9\sqrt{2} = \sqrt{81 \times 2} = \sqrt{162}$. Note the options are under the square root — don't forget to square back when matching the answer.

Given: $a_n = 20 + 4n$

(i) Number on the 1st spot:

$a_1 = 20 + 4(1) = 20 + 4 = \mathbf{24}$

(ii) OR Option 1: Which spot is numbered 112?

$20 + 4n = 112$
$4n = 92 \Rightarrow n = 23$

∴ 112 is on the 23rd spot.

OR Option 2: Sum of first 10 terms:

$a_1 = 24,\ a_{10} = 20 + 4(10) = 60$

$S_{10} = \dfrac{10}{2}(a_1 + a_{10}) = 5 \times (24 + 60) = 5 \times 84 = \mathbf{420}$

(iii) Number on the $(n-2)$th spot:

Replace $n$ with $(n-2)$:

$a_{n-2} = 20 + 4(n-2) = 20 + 4n - 8 = \mathbf{4n + 12}$

---

• Part (i): Simply substitute $n = 1$ in the given formula.
• Part (ii): Set $a_n = 112$ and solve for $n$ (Option 1), or use the sum formula $S_n = \frac{n}{2}(a_1 + a_n)$ (Option 2). Attempt only ONE option in the exam.
• Part (iii): Substitute $(n-2)$ in place of $n$ in the formula and simplify — a common 1-mark algebraic substitution step.

Option B: 10

Let the three numbers in A.P. be $a-d,\ a,\ a+d$. Their sum $= 3a = 30$, so $a = 10$. The middle term is 10.

The standard trick for three numbers in A.P. is to take them as $a-d, a, a+d$. The $d$ cancels on adding, giving $3a =$ sum directly. The middle term equals $\frac{\text{sum}}{3} = \frac{30}{3} = 10$.

Given: $\dfrac{a_{10}}{a_{30}} = \dfrac{1}{3}$ and $S_6 = 42$

Setting up equations:

$$a_{10} = a + 9d, \quad a_{30} = a + 29d$$

$$\frac{a + 9d}{a + 29d} = \frac{1}{3}$$

$$3a + 27d = a + 29d$$

$$2a = 2d \implies a = d \quad \text{...(1)}$$

For $S_6 = 42$:

$$\frac{6}{2}[2a + 5d] = 42$$

$$3[2a + 5d] = 42 \implies 2a + 5d = 14 \quad \text{...(2)}$$

Substituting (1) into (2):

$$2d + 5d = 14 \implies 7d = 14 \implies d = 2$$

$$a = 2$$

First term $a = 2$, common difference $d = 2$.

Source: Arithmetic Progressions, Chapter 5

---

• Use $a_n = a + (n-1)d$ to form the ratio equation, cross-multiply, and simplify to get the first relation between $a$ and $d$.
• Use the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ for the second equation.
• Solve the two simultaneous equations. Here both give $a = d = 2$, so verify by substitution.
• Show all steps clearly — each equation and substitution carries marks.

Given: $S_7 = 49$ and $S_{17} = 289$

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

$$S_7 = \frac{7}{2}[2a + 6d] = 49 \implies 2a + 6d = 14 \implies a + 3d = 7 \quad \text{...(1)}$$

$$S_{17} = \frac{17}{2}[2a + 16d] = 289 \implies 2a + 16d = 34 \implies a + 8d = 17 \quad \text{...(2)}$$

Subtracting (1) from (2): $5d = 10 \implies d = 2$

From (1): $a + 6 = 7 \implies a = 1$

$$S_{20} = \frac{20}{2}[2(1) + 19(2)] = 10[2 + 38] = 10 \times 40 = \boxed{400}$$

Source: Chapter 5, Exercise 5.3 (Q9)

---

• Set up two equations using the $S_n$ formula for $n = 7$ and $n = 17$, then solve simultaneously for $a$ and $d$.
• Examiners award marks for: forming the two equations (1 mark), solving to get $a = 1, d = 2$ (1 mark), and correct $S_{20}$ (1 mark).
• Note: The textbook Q9 asks for $S_n$ in general; this question specifically asks for $S_{20}$, so substitute $n = 20$ at the end.

$d = a_2 - a_1 = \dfrac{1-4x}{2x} - \dfrac{1}{2x} = \dfrac{1-4x-1}{2x} = \dfrac{-4x}{2x} = -2$

The correct option is (B) $-2$.

Find $d$ by subtracting the first term from the second term. Both terms share the denominator $2x$, so subtract only the numerators: $(1-4x) - 1 = -4x$, giving $\frac{-4x}{2x} = -2$. The $x$ cancels, so $d$ is a constant $-2$, not involving $x$.

Let the three consecutive terms of the A.P. be $(a-d)$, $a$, $(a+d)$.

Condition 1: Sum = 24
$$(a-d) + a + (a+d) = 24$$
$$3a = 24 \implies a = 8$$

Condition 2: Sum of squares = 194
$$(a-d)^2 + a^2 + (a+d)^2 = 194$$
$$3a^2 + 2d^2 = 194$$
$$3(64) + 2d^2 = 194$$
$$2d^2 = 194 - 192 = 2 \implies d^2 = 1 \implies d = \pm 1$$

When $d = 1$: terms are 7, 8, 9

When $d = -1$: terms are 9, 8, 7

∴ The three numbers are 7, 8, 9.

Source: Chapter 5, Arithmetic Progressions

---

• Always assume consecutive A.P. terms as $(a-d), a, (a+d)$ — this makes the sum condition simple (middle term directly gives $a$).
• Expand $(a-d)^2 + a^2 + (a+d)^2 = 3a^2 + 2d^2$ — a standard identity to remember.
• Both $d = +1$ and $d = -1$ give the same set of numbers, so either answer is accepted; state both for full marks.

Given: $S_m = S_n$, where $m \neq n$

Using the formula $S_n = \dfrac{n}{2}[2a + (n-1)d]$:

$$\frac{m}{2}[2a + (m-1)d] = \frac{n}{2}[2a + (n-1)d]$$

$$m[2a + (m-1)d] - n[2a + (n-1)d] = 0$$

$$2a(m-n) + d[m(m-1) - n(n-1)] = 0$$

$$2a(m-n) + d[(m^2 - n^2) - (m - n)] = 0$$

$$2a(m-n) + d(m-n)[(m+n) - 1] = 0$$

Since $m \neq n$, divide by $(m - n)$:

$$2a + (m+n-1)d = 0 \quad \cdots (1)$$

Now, sum of first $(m+n)$ terms:

$$S_{m+n} = \frac{(m+n)}{2}[2a + (m+n-1)d] = \frac{(m+n)}{2} \times 0 = \boxed{0}$$

[From equation (1)]

Hence proved.

Source: Chapter 5, Section 5.4 (Sum of first $n$ terms of an AP)

---

• The key step is subtracting the two equal sum expressions and factoring out $(m-n)$, which is non-zero, giving $2a + (m+n-1)d = 0$.
• Examiners award marks for: setting up $S_m = S_n$ correctly (1 mark), reaching $2a+(m+n-1)d=0$ (1 mark), and concluding $S_{m+n}=0$ (1 mark).
• Notice that $2a + (m+n-1)d$ is exactly the bracket in the formula for $S_{m+n}$—that's the elegant connection to spot.

(B) 16th

Here, $a = -29$, $d = 3$, $a_n = 16$.
Using $a_n = a + (n-1)d$: $16 = -29 + (n-1) \times 3 \Rightarrow 45 = 3(n-1) \Rightarrow n = 16$.

Source: Chapter 5, Section 5.3 (nth term formula)

---

Use the formula $a_n = a + (n-1)d$. Here $a = -29$, $d = -26-(-29) = 3$, and $a_n = 16$. Solving gives $n = 16$. A common mistake is arithmetic error when dividing; always verify by substituting $n$ back.

(A) 12

$a_{15} - a_{11} = [a+14d] - [a+10d] = 4d = 48 \Rightarrow d = 12$

Use $a_n = a+(n-1)d$. Subtracting the two terms eliminates $a$, leaving $4d = 48$, so $d = 12$. Only option (A) matches.

Given: First term $a = 8$ years, common difference $d = 4$ months $= \dfrac{1}{3}$ year, Sum $S_n = 168$ years.

Using the formula:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

$$168 = \frac{n}{2}\left[2 \times 8 + (n-1) \times \frac{1}{3}\right]$$

$$168 = \frac{n}{2}\left[16 + \frac{n-1}{3}\right]$$

$$336 = n\left[\frac{48 + n - 1}{3}\right]$$

$$1008 = n(47 + n)$$

$$n^2 + 47n - 1008 = 0$$

$$n^2 + 63n - 16n - 1008 = 0$$

$$(n + 63)(n - 16) = 0$$

Since $n$ cannot be negative, $n = 16$.

Age of the eldest participant:

$$a_{16} = a + (n-1)d = 8 + 15 \times \frac{1}{3} = 8 + 5 = 13 \text{ years}$$

The age of the eldest participant is 13 years.

Source: Arithmetic Progressions, Section 5.4

---

• Convert $d = 4$ months to years ($\frac{1}{3}$ year) before substituting — a common error point.
• Substitute into $S_n = \frac{n}{2}[2a + (n-1)d]$, simplify to get a quadratic, and reject the negative root.
• Finally use $a_n = a + (n-1)d$ to find the last term. Examiners award marks at each step: identifying AP values (1 mark), setting up the sum equation (1 mark), solving the quadratic (2 marks), and finding the eldest age (1 mark).

Let the first term be $a$ and common difference be $d$.

Setting up equations:

$a_3 = a + 2d$ and $a_7 = a + 6d$

Given: $a_3 + a_7 = 6$
$$( a + 2d) + (a + 6d) = 6 \implies 2a + 8d = 6 \implies a + 4d = 3 \quad \text{...(1)}$$

Given: $a_3 \times a_7 = 8$
$$(a + 2d)(a + 6d) = 8 \quad \text{...(2)}$$

Solving:

From (1): $a = 3 - 4d$. Substituting in (2):

$$(3 - 4d + 2d)(3 - 4d + 6d) = 8$$
$$(3 - 2d)(3 + 2d) = 8$$
$$9 - 4d^2 = 8 \implies 4d^2 = 1 \implies d = \pm\frac{1}{2}$$

Case 1: $d = \frac{1}{2}$, then $a = 3 - 4\times\frac{1}{2} = 1$

Case 2: $d = -\frac{1}{2}$, then $a = 3 - 4\times\left(-\frac{1}{2}\right) = 5$

Sum of first 16 terms:

$$S_{16} = \frac{16}{2}[2a + 15d] = 8[2a + 15d]$$

Case 1: $S_{16} = 8\left[2(1) + 15\times\frac{1}{2}\right] = 8\left[2 + 7.5\right] = 8 \times 9.5 = \mathbf{76}$

Case 2: $S_{16} = 8\left[2(5) + 15\times\left(-\frac{1}{2}\right)\right] = 8\left[10 - 7.5\right] = 8 \times 2.5 = \mathbf{20}$

∴ The sum of the first 16 terms is 76 or 20.

Source: Chapter 5, Exercise 5.4 Q.2

---

• The key trick is using the difference of squares identity: $(3-2d)(3+2d) = 9 - 4d^2$, which quickly gives $d$.
• Since $d$ has two values, there are two valid APs — present both cases for full marks.
• Examiners award marks at each step: setting up equations (1 mark), solving for $d$ and $a$ (2 marks), and computing $S_{16}$ for both cases (2 marks).
• Do not forget the formula $S_n = \frac{n}{2}[2a+(n-1)d]$.

Simple Interest formula: $\text{SI} = \frac{P \times R \times T}{100}$

Interest at end of 1st year $= \frac{2000 \times 7 \times 1}{100} = ₹140$

Interest at end of 2nd year $= \frac{2000 \times 7 \times 2}{100} = ₹280$

Interest at end of 3rd year $= \frac{2000 \times 7 \times 3}{100} = ₹420$

The interests are 140, 280, 420, …

Since $280 - 140 = 420 - 280 = 140$, the common difference is constant. Yes, they form an AP with $a = 140$ and $d = 140$.

Interest at end of 27th year:

$$a_{27} = a + (27-1)d = 140 + 26 \times 140 = 140 \times 27 = ₹3780$$

Source: Chapter 5, Section 5.3 (Example 9 pattern)

---

• This question is modelled exactly on Example 9 of Section 5.3, but with P = ₹2000, R = 7% instead of P = ₹1000, R = 8%.
• Examiners expect you to: (1) calculate the three interest values, (2) explicitly verify the AP by checking equal differences, and (3) apply $a_n = a + (n-1)d$ for the 27th term.
• Note: $a_{27} = 140 \times 27$ is a neat shortcut since $a = d = 140$.
• Do not confuse "interest at end of $n$th year" with total amount — it is only the SI for that cumulative time.

(D) Assertion (A) is false, but Reason (R) is true.

The common difference of AP: 5, 1, −3, … is $d = 1 − 5 = −4$, not 4. The Reason is correct as $d = a_n − a_{n−1}$.

Source: Chapter 5, Section 5.1 / Summary Point 2

• The Assertion gives $d = 4$, which is wrong; the correct value is $d = -4$ (each term decreases by 4).
• The Reason correctly states the formula for common difference, so R is true.
• Since A is false but R is true, option (D) is correct.
• Always verify the sign when calculating common difference — a common mistake is ignoring the negative sign.

Here, $a = -5$, $d = -\dfrac{5}{2} - (-5) = \dfrac{5}{2}$, $n = 10$.

$a_{10} = a + 9d = -5 + 9 \times \dfrac{5}{2} = -5 + \dfrac{45}{2} = \dfrac{35}{2}$

Answer: (A) $\dfrac{35}{2}$

---

Use the nth term formula $a_n = a + (n-1)d$ from the chapter summary. Identify $a = -5$ and $d = \frac{5}{2}$ (positive, since the sequence increases). Substituting $n = 10$ gives $\frac{35}{2}$. A common mistake is taking $d$ as negative — check the sign carefully from consecutive terms.

The AP formed is: 300, 350, 400, 450, 500, 550, …
Here, first term $a = 300$ and common difference $d = 50$.

(i) 4th term = 450 m, 5th term = 500 m, 6th term = 550 m

(ii) Distance of 8th round:
$a_8 = a + 7d = 300 + 7 \times 50 = 300 + 350 = \mathbf{650}$ metres

(iii) Total distance after 10 rounds:
$$S_{10} = \frac{n}{2}[2a + (n-1)d] = \frac{10}{2}[2 \times 300 + 9 \times 50]$$
$$= 5 \times [600 + 450] = 5 \times 1050 = \mathbf{5250 \text{ metres}}$$

Source: Arithmetic Progressions, AP Sum Formula

---

• Examiner focus: Correct identification of $a$ and $d$ is key; errors here lose marks throughout.
• (i) Simply list terms — no formula needed.
• (ii) Use $a_n = a + (n-1)d$; substitution must be shown for full credit.
• (iii) Must write the $S_n$ formula, substitute values, and show working step-by-step — both steps carry marks. Final answer in metres.

$S_m = 2m^2 + 3m$

$a_2 = S_2 - S_1 = (2(4)+3(2)) - (2(1)+3(1)) = 14 - 5 = \mathbf{9}$

Answer: (B) 9

The nth term of an AP can be found using $a_n = S_n - S_{n-1}$. Substitute $m=2$ and $m=1$ into the given sum formula and subtract. Avoid confusing $S_n$ (sum) with $a_n$ (term) — a common mistake.

(i)
The number of dots: 4, 8, 12, …

Differences: 8 − 4 = 4, 12 − 8 = 4 (constant)

Since the common difference is constant, the dots form an A.P.
First term (a) = 4, Common difference (d) = 4

---

(ii)
$a_n = a + (n-1)d = 4 + (n-1) \times 4 = 4n$

$$\boxed{a_n = 4n}$$

---

(iii)
Given: $S_n = 220$

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

$$220 = \frac{n}{2}[8 + 4(n-1)]$$

$$440 = n[4n + 4] = 4n(n+1)$$

$$n(n+1) = 110$$

$$n^2 + n - 110 = 0$$

$$(n-10)(n+11) = 0$$

$$n = 10 \quad (\text{rejecting } n = -11)$$

∴ 10 squares are formed.

Source: Arithmetic Progressions, CBSE Class 10 Mathematics

---

• (i) Just show the differences are equal — that's enough for 1 mark. State a and d clearly.
• (ii) Substitute the formula directly; simplify to 4n. Examiners expect the simplified form.
• (iii) Use the sum formula (not nth term). Set up the quadratic, solve, and reject the negative root — both steps carry marks. Always state why n = −11 is rejected.

Option A: $\sqrt{2}$

$d = a_{n+1} - a_n = [\sqrt{2}(n+1)+1] - [\sqrt{2}n+1] = \sqrt{2}$

The common difference is found by subtracting the $n$th term from the $(n+1)$th term. Since $\sqrt{2}\,n$ cancels and $+1$ cancels, only $\sqrt{2}$ remains. Note that $d$ must be a constant (no $n$ in it), which immediately rules out option B.