Arithmetic Progressions

Class 10 Mathematics � Chapter 05 (Deep Detail)

1. Fundamental Terms

Arithmetic Progression (AP): A sequence where difference between consecutive terms is constant. $a, a+d, a+2d \dots$

[AI IMAGE PROMPT: A 3D illustration of a staircase. Each step is exactly the same height 'd'. A character climbs from 'a' (Step 1) to 'a+d' (Step 2) to 'a+2d' (Step 3). Labels showing the steady linear progression.]

Term	Symbol	Formula/Note
First Term	$a$	Starting number of the sequence.
Common Difference	$d$	$d = a_2 - a_1 = a_3 - a_2$. Can be +ve, -ve, or 0.
nth Term	$a_n$	$a_n = a + (n-1)d$
Sum of n terms	$S_n$	$S_n = \frac{n}{2}[2a + (n-1)d]$

[AI IMAGE PROMPT: Visualizing Sum of AP ($S_n$). Stacks of blocks arranged in increasing order (like mobile signal bars) forming a triangle. A mirrored copy of the triangle is placed upside down on top, forming a perfect Rectangle. Formula $S_n = \frac{n(n+1)}{2}$ visualized as Area of half-rectangle.]

2. Advanced Tricks for Unknown Terms

When Sum of terms is given in question, assume terms symmetrically to cancel 'd'.

3 Terms: $a-d, \quad a, \quad a+d$. (Sum = $3a$)
4 Terms: $a-3d, \quad a-d, \quad a+d, \quad a+3d$. (Sum = $4a$, Common Diff = 2d)
5 Terms: $a-2d, \quad a-d, \quad a, \quad a+d, \quad a+2d$. (Sum = $5a$)

Example: 3 Angles of Triangle in AP

Q: Angles of a triangle are in AP. The greatest is twice the least. Find angles.

Let angles be $a-d, a, a+d$.
Sum Property: $(a-d) + a + (a+d) = 180^\circ \Rightarrow 3a = 180^\circ \Rightarrow a = 60^\circ$.
Greatest ($60+d$) = 2 $\times$ Least ($60-d$).
$60+d = 120 - 2d \Rightarrow 3d = 60 \Rightarrow d = 20^\circ$.
Angles: $40^\circ, 60^\circ, 80^\circ$.

3. Relationship between $S_n$ and $a_n$

GOLDEN RELATION

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

(nth term is the difference between Sum of n terms and Sum of n-1 terms)

Example: Finding AP from Sum Formula

Q: If sum of first n terms is $S_n = 3n^2 + n$, find the AP.

Find $a_1$: $S_1 = 3(1)^2 + 1 = 4$. So, first term = 4.
Find $S_2$: $S_2 = 3(2)^2 + 2 = 12 + 2 = 14$. (Sum of first 2 terms).
Find $a_2$: $a_2 = S_2 - S_1 = 14 - 4 = 10$.
Find d: $d = a_2 - a_1 = 10 - 4 = 6$.
AP: 4, 10, 16, 22...

4. High-Level Word Problems (Case Studies)

LADDER PROBLEM A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at bottom to 25 cm at top. Top and bottom rungs are $2 \frac{1}{2}$ m apart. What is length of wood required?

[AI IMAGE PROMPT: A wooden ladder leaning against a wall. The bottom rung is wide (45cm). The rungs get narrower and narrower as they go up, ending at the top rung (25cm). Spacing between rungs is marked "25 cm". Total height marked "2.5 m".]

Number of gaps: $250 \text{ cm} / 25 \text{ cm} = 10$.
Number of rungs (n): Gaps + 1 = $11$ rungs. (Crucial Step!)
$a = 25$ (top), $l = 45$ (bottom).
$S_{11} = \frac{11}{2} [25 + 45] = \frac{11}{2} [70] = 385$ cm.

Potato Race Problem (Standard Case Study)

Scenario: A bucket is at starting point. First potato is 5m away. Other potatoes are 3m apart. There are 10 potatoes. Competitor starts from bucket, picks nearest potato, runs back to drop it, runs for second, etc.

[AI IMAGE PROMPT: A fun sports field layout for "Potato Race". A bucket is at the start line 0m. A potato lies at 5m. Another at 8m, 11m, etc. A runner is shown sprinting back and forth. Arrows show the path: Start -> Potato 1 -> Bucket -> Potato 2 -> Bucket. Text label: "Distance Doubles Each Try".]

Distance Calculation:

For Potato 1: Run 5m, back 5m. Total = 10m.
For Potato 2: Run $5+3=8$m, back 8m. Total = 16m.
For Potato 3: Run $8+3=11$m, back 11m. Total = 22m.
AP: 10, 16, 22... ($a=10, d=6$).
Total Distance = $S_{10} = \frac{10}{2} [2(10) + (9)(6)] = 5 [20 + 54] = 5 \times 74 = 370$ m.

5. Arithmetic Mean

If $a, b, c$ are in AP, then $b$ is the Arithmetic Mean of $a$ and $c$.

$$ 2b = a + c \quad \text{or} \quad b = \frac{a+c}{2} $$