Arithmetic Progressions | Class 10 Maths

🌟 AP Kya Hota Hai?

🔍 Real Life mein AP:

• Bus stop par numbers: 1, 2, 3, 4, 5... (d=1)
• EMI payment: ₹500, ₹500, ₹500... (d=0)
• Saved money: ₹100, ₹200, ₹300... (d=100)
• Countdown: 10, 7, 4, 1, −2... (d=−3)
Sab mein common — ek fixed difference!

Arithmetic Progression (AP): Number sequence jisme consecutive terms ka difference constant hota hai.
Ye constant difference = Common Difference (d)
d = a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = ...

AP Sequence Visualise Karo:

a +d→ a+d +d→ a+2d +d→ a+3d +d→ ...

Sequence	First term (a)	Common diff (d)	AP?
2, 4, 6, 8, 10	2	2	✅
3, 7, 11, 15	3	4	✅
1, 4, 9, 16	1	3,5,7... (changing)	❌
5, 5, 5, 5	5	0	✅ (constant)
10, 7, 4, 1, −2	10	−3	✅ (decreasing)

📐 nth Term Formula (aₙ)

Observe: a₁=a, a₂=a+d, a₃=a+2d, ...
Pattern: aₙ = a + (n−1)d

aₙ = a + (n−1)d

Alternative: agar last term 'l' pata ho, to: aₙ = l − (n−1)d (from end).
Also: l = a + (n−1)d jahan l = last term

Example 1: AP: 5, 8, 11, 14... ka 20th term nikalo.

a=5, d=3. a₂₀ = 5 + 19×3 = 5 + 57 = 62

Example 2: Kaunsa term AP (3, 8, 13, 18...) mein 78 hai?

a=3, d=5. aₙ = 78 → 3 + (n−1)5 = 78 → (n−1)5 = 75 → n−1 = 15 → n = 16

Example 3: AP ka 3rd term 16, 7th term 28. AP find karo.

a₃ = a+2d = 16 ...(1)
a₇ = a+6d = 28 ...(2)
(2)−(1): 4d = 12 → d = 3
From (1): a = 16−6 = 10
AP: 10, 13, 16, 19, 22, 25, 28...

Example 4: 11th term se 13th term? AP: 3, 15, 27, 39...

d = 12. a₁₁ = 3 + 10×12 = 123. a₁₃ = 3 + 12×12 = 147.
Difference = 147−123 = 24 = 2d ✓

📊 Sum of n Terms (Sₙ)

🌟 Gauss ki Kahani (9 saal ki umar mein!)

Teacher ne class ko 1 se 100 tak ke numbers jodne ko kaha — time waste karne ke liye. Lekin Carl Friedrich Gauss ne sirf kuch seconds mein answer diya!

Trick:
S = 1+2+3+...+98+99+100
S = 100+99+98+...+3+2+1
─────────────────────────
2S = 101+101+...+101 (100 baar) = 10100
S = 5050 ✓

Isi Gauss trick se general formula:

When last term known

Sₙ = n/2 (a + l)

l = last term

When d is known

Sₙ = n/2 [2a + (n−1)d]

Both formulas same hain!

Example 1: 1+2+3+...+50 = ?

a=1, l=50, n=50. S = 50/2 × (1+50) = 25 × 51 = 1275

Example 2: AP: 5, 7, 9... ke 25 terms ka sum

a=5, d=2, n=25. S₂₅ = 25/2 × [10 + 24×2] = 25/2 × 58 = 25×29 = 725

Example 3: Sum of first n natural numbers = ?

a=1, d=1. Sₙ = n/2[2 + (n−1)] = n(n+1)/2
S₁₀₀ = 100×101/2 = 5050 (Gauss ka answer!)

Example 4: S₇ = 49 aur a₁₇ = 5(a₇). AP find karo.

S₇ = 7/2 (2a+6d) = 7(a+3d) = 49 → a+3d = 7 ...(1)
a₁₇ = a+16d = 5(a+6d) = 5a+30d → −4a−14d=0 → 2a+7d=0 ...(2)
From (1): a=7−3d. In (2): 14−6d+7d=0 → 14+d=0 → d=−14. a=7+42=49.
AP: 49, 35, 21, 7, −7...

🔍 Important Relations

aₙ = Sₙ − Sₙ₋₁ (nth term = difference of consecutive sums)
Sum of odd terms: 1+3+5+...+(2n−1) = n²
Sum of even terms: 2+4+6+...+2n = n(n+1)

Example: Agar Sₙ = 3n² + 4n, to a₁ aur a₂ nikalo.

S₁ = 3+4 = 7 = a₁ (kyunki S₁ = a₁)
S₂ = 12+8 = 20 = a₁+a₂ → a₂ = 20−7 = 13
d = a₂ − a₁ = 13 − 7 = 6 (verify: aₙ = Sₙ−Sₙ₋₁ = 6n+1, a₁=7 ✓)

📝 Practice Questions

Q1. AP: 7, 13, 19... ka 30th term kya hai?

a=7, d=6. a₃₀ = 7+29×6 = 7+174 = 181

Q2. 1+3+5+7+...+199 = ?

Odd numbers. n = (199+1)/2 = 100. Sum of first 100 odd numbers = 100² = 10000

Q3. AP ka 4th term 22 hai aur common difference −4 hai. Pehle 10 terms ka sum nikalo.

a₄=a+3d=22 → a=22+12=34. S₁₀ = 10/2[68+9×(−4)] = 5[68−36] = 5×32 = 160

Q4. Ek AP mein 1st term = 5, last term = 45, sum = 400. Terms ki sankhya aur common difference nikalo.

S = n/2(5+45)=400 → n×25=400 → n=16. d=(45−5)/15 = 40/15 = 8/3

Q5. Kaunse AP mein 5th aur 9th terms ka ratio 5:8 hai aur sum of 8 terms = 136?

a₅/a₉ = (a+4d)/(a+8d) = 5/8 → 8a+32d=5a+40d → 3a=8d ...(1)
S₈ = 4(2a+7d) = 136 → 2a+7d=34 ...(2)
From (1): a=8d/3. In (2): 16d/3+7d=34 → 37d/3=34 → d=102/37... Hmm let me try integer values.
Actually: 3a=8d → a=8, d=3 satisfies (let's verify: 2(8)+7(3)=16+21=37≠34).
Let a=8d/3 in (2): 16d/3+7d=34 → 37d=102 → d=102/37. Try d=3: a=8. 2(8)+7(3)=37. No.
AP: a=8d/3, d=3 → a=8. Sequence: 8,11,14,17,20,23,26,29 (S₈=8+11+...+29=148≠136. So let's recheck)
Correct: d=3, a=8: S₈=4(16+21)=4×37=148. Ratio: a₅/a₉=20/32=5/8 ✓ but S≠136.
Try S₈=136 → 2a+7d=34 with 3a=8d: a=8d/3 → 16d/3+7d=34 → d=102/37 (non-integer).
Note: Some problems have fractional answers — ye valid hai! d=102/37, a=272/37.

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