1. Arithmetic progression (A.P.)
Each term = previous + common difference d.
- nth term: Tₙ = a + (n − 1)d
- Sum: Sₙ = n/2 [2a + (n − 1)d] = n/2 (first + last)
- Tₙ = Sₙ − Sₙ₋₁; first term a = S₁.
- Three terms in A.P.: a − d, a, a + d. Middle term = average of neighbours: 2b = a + c.
- nth term from the end = l − (n − 1)d.
2. Special sums
| Sum | Formula |
|---|---|
| 1 + 2 + … + n | n(n + 1)/2 |
| First n odd numbers | n² |
| First n even numbers | n(n + 1) |
| 1² + … + n² | n(n + 1)(2n + 1)/6 |
| 1³ + … + n³ | [n(n + 1)/2]² |
3. A.P. solved examples
(ASO 2021) T₇ = 34, T₁₃ = 64 → 6d = 30, d = 5, a = 4 → T₁₈ = 89.
(ASO 2021) Sₙ = 3n² + 5n → Tₙ = 6n + 2 = 164 → n = 27.
(ASO 2012) 63, 65, 67 … and 3, 10, 17 … have equal nth terms → 63 + 2(n − 1) = 3 + 7(n − 1) → n = 13.
(ASO 2015) Is 301 in 5, 11, 17 …? 5 + 6(n − 1) = 301 → n not whole → No.
Numbers between 100 and 300 divisible by 7: 105 … 294 → (294 − 105)/7 + 1 = 28.
4. Geometric progression (G.P.)
Each term = previous × common ratio r.
- Tₙ = a rⁿ⁻¹
- Sₙ = a(rⁿ − 1)/(r − 1) for r > 1; a(1 − rⁿ)/(1 − r) for r < 1
- Infinite sum (|r| < 1): a/(1 − r)
- Geometric mean of a, b = √ab; b² = ac for three terms in G.P.
3, 6, 12 … T₈ = 3 × 2⁷ = 384; S₆ = 3(64 − 1)/1 = 189.
1 + 1/2 + 1/4 + … = 1/(1 − 1/2) = 2.
(ASO 2021) GM of 4 and 16 = √64 = 8.
Solved examples by question type
Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.
Basic level
Example 1. The 25th term of the AP 2, 7, 12, … is:
(a) 128 (b) 127 (c) 122 (d) 117
Answer: (c) 122
Solution: aₙ = a + (n−1)d = 2 + 24×5 = 122.
Example 2. The 5th term of the G.P. 3, 12, 48, … is:
(a) 192 (b) 19 (c) 768 (d) 3,072
Answer: (c) 768
Solution: Tn = a r^(n−1) = 3 × 4^4 = 768.
Example 3. The sum of the first 40 even natural numbers is:
(a) 820 (b) 1,600 (c) 1,640 (d) 1,680
Answer: (c) 1,640
Solution: Sum of first n even numbers = n(n + 1) = 1,640.
ASO exam level
Example 4. If the 3th and 8th terms of an AP are 12 and 27, then its 15th term is:
(a) 34 (b) 45 (c) 48 (d) 51
Answer: (c) 48
Solution: d = (27 − 12)/(8 − 3) = 3. a = 12 − 2×3 = 6. 15th term = 6 + 14×3 = 48.
Example 5. Find the sum of the first 25 terms of the AP 1, 4, 7, …
(a) 925 (b) 1,000 (c) 1,825 (d) 924
Answer: (a) 925
Solution: Sₙ = n/2 [2a + (n−1)d] = 25/2 × [2 + 72] = 925.
Example 6. Find the sum of the first 20 terms of the AP 3, 7, 11, …
(a) 900 (b) 820 (c) 817 (d) 1,580
Answer: (b) 820
Solution: Sₙ = n/2 [2a + (n−1)d] = 20/2 × [6 + 76] = 820.
Example 7. Which term of the AP 2, 6, 10, … is 106?
(a) 29th (b) 26th (c) 28th (d) 27th
Answer: (d) 27th
Solution: 2 + (n−1)×4 = 106 → n = 27.
Example 8. The sum of the first 7 terms of the G.P. 4, 12, 36, … is:
(a) 4,372 (b) 13,120 (c) 4,368 (d) 2,916
Answer: (a) 4,372
Solution: S = a(rⁿ − 1)/(r − 1) = 4(3^7 − 1)/2 = 4,372.
Example 9. The sum of the first 6 terms of the G.P. 3, 9, 27, … is:
(a) 1,092 (b) 729 (c) 1,089 (d) 3,279
Answer: (a) 1,092
Solution: S = a(rⁿ − 1)/(r − 1) = 3(3^6 − 1)/2 = 1,092.
Example 10. How many numbers between 128 and 921 are divisible by 13?
(a) 62 (b) 63 (c) 60 (d) 61
Answer: (d) 61
Solution: First = 130, last = 910. Count = (910 − 130)/13 + 1 = 61.
Advanced level
Example 11. The sum of the first 15 terms of the A.P. 5, 7, 9, … is:
(a) 315 (b) 285 (c) 495 (d) 280
Answer: (b) 285
Solution: S = n/2 [2a + (n − 1)d] = 15/2 [10 + 28] = 285.
Example 12. The sum of the first 11 terms of the A.P. 14, 20, 26, … is:
(a) 550 (b) 814 (c) 470 (d) 484
Answer: (d) 484
Solution: S = n/2 [2a + (n − 1)d] = 11/2 [28 + 60] = 484.
Topic-wise practice for Arithmetic and geometric progression (Quant): nth term, sum, means, sequences. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.
Question types covered
- nth term of an AP
- Sum of n terms of an AP
- Three or four numbers in AP
- Which term is a given number
- nth term and sum of a GP
- Arithmetic and geometric mean
- Finding the series from Sn