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Chapter 5: Arithmetic Progressions

Board Exam Focused Notes, Formulas, and PYQs

Exam Weightage & Blueprint

Total: 6-7 Marks

This chapter is part of the Algebra Unit. It's a high-scoring chapter with formula-based questions.

Question Type Marks Frequency Focus Topic
MCQ 1 High Finding 'd', 'a', or nth term
Short Answer 2 or 3 Medium Sum of AP ($S_n$), finding $n$
Case Study 4 High Real-life Applications (Ladders, Savings)

⏰ Last 24-Hour Checklist

  • Definition: Fixed difference between terms.
  • Common Difference (d): $a_2 - a_1$.
  • General Form: $a, a+d, a+2d...$
  • nth Term ($a_n$): $a + (n-1)d$.
  • Sum ($S_n$): $\frac{n}{2}[2a+(n-1)d]$.
  • nth Term from Sum: $a_n = S_n - S_{n-1}$.

Concepts & Definitions ★★★★★

Arithmetic Progression (AP): A list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

Key Components

First Term (a)

The starting number of the sequence.

Difference (d)

$d = a_{k+1} - a_k$. Can be $+$, $-$, or $0$.

General Term ($a_n$)

Value of the term at position $n$.

Check for AP: To check if a list is an AP, calculate differences ($a_2-a_1$, $a_3-a_2$). If they are EQUAL, it's an AP.

Important Formulas 🔥🔥🔥

1. General Term

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

2. nth Term from the END (Exam Secret)

If $l$ is the last term, use this direct formula instead of reversing the AP:

$$ \text{nth term from end} = l - (n - 1)d $$

3. Sum of First n Terms ($S_n$)

$$ S_n = \frac{n}{2} [ 2a + (n - 1)d ] $$
OR (if last term $l$ is known)
$$ S_n = \frac{n}{2} ( a + l ) $$

4. Finding Term from Sum (HOTS)

If $S_n$ is given (e.g., $S_n = 4n - n^2$) and you need to find the AP or $a_n$:

$$ a_n = S_n - S_{n-1} \quad (\text{for } n > 1) $$ $$ a_1 = S_1 $$

Solved Examples (Board Marking Scheme)

Q1. Find the 11th term from the last term of AP: 10, 7, 4, ..., -62. (2 Marks)

Step 1: Identify Parameters 0.5 Mark

$a=10$, $d = 7-10 = -3$, Last term $l = -62$, $n=11$.

Step 2: Apply Formula 1 Mark

nth term from end $= l - (n-1)d$

$= -62 - (11-1)(-3)$

$= -62 - (10)(-3)$

Step 3: Calculate 0.5 Mark

$= -62 + 30 = -32$.

Q2. If the sum of first n terms is $S_n = 4n - n^2$, find the 10th term. (3 Marks)

Step 1: Find S10 and S9 1.5 Marks

$S_{10} = 4(10) - (10)^2 = 40 - 100 = -60$.

$S_9 = 4(9) - (9)^2 = 36 - 81 = -45$.

Step 2: Apply Formula 1 Mark

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

$a_{10} = S_{10} - S_9$

$a_{10} = -60 - (-45)$

Step 3: Answer 0.5 Mark

$a_{10} = -60 + 45 = -15$.

Q3. Which term of the AP: 21, 18, 15, ... is -81? (3 Marks)

Step 1: Identify Parameters 0.5 Mark

$a = 21$, $d = 18 - 21 = -3$, $a_n = -81$.

Step 2: Apply Formula 1 Mark

$-81 = 21 + (n-1)(-3)$

$-102 = -3(n-1)$

Step 3: Solve 1.5 Marks

$34 = n - 1 \Rightarrow n = 35$.

Therefore, the 35th term is -81.

Previous Year Questions (PYQs)

2023 (1 Mark): Find the 10th term of AP: 2, 7, 12...
Ans: $a=2, d=5, n=10$. $a_{10} = 2 + 9(5) = 47$.
2020 (3 Marks): How many two-digit numbers are divisible by 3?
Ans: Sequence: 12, 15, ..., 99. Here $a=12, d=3, a_n=99$. Solving $99=12+(n-1)3$ gives $n=30$.
2019 (4 Marks): The sum of 4th and 8th terms of an AP is 24 and sum of 6th and 10th terms is 44. Find the AP.
Ans: Eq1: $a+3d + a+7d = 24 \Rightarrow 2a+10d=24$. Eq2: $2a+14d=44$. Solving gives $d=5, a=-13$. AP: -13, -8, -3...

Exam Strategy & Mistake Bank

Mistake Bank 🚨

Common Difference Sign: If AP is decreasing (10, 8, 6...), $d$ must be negative ($-2$). Students often write $+2$.
n vs nth term: Confusing "number of terms" ($n$) with "value of term" ($a_n$).
Calculation: In $S_n$ formula, forgetting to multiply $n/2$ with the bracket content.

Scoring Tips 🏆

Write Formula: Always write the generic formula ($a_n = ...$) before substituting values. It carries marks.
Case Studies: For word problems (flower beds, savings), list the sequence first (e.g., 20, 25, 30...) then apply AP formulas.
n must be integer: If you calculate $n$ as a fraction (e.g., 15.5), recheck your work. Number of terms cannot be decimal.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): Write the common difference of AP: $3, 1, -1, -3...$

Q2 (2M): Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Q3 (3M): How many terms of the AP: 9, 17, 25... must be taken to give a sum of 636?

Q4 (4M): If the sum of first $n$ terms is $4n - n^2$, find the first term and the 2nd term.

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