Progressions Series
Quick Theory
A Progression is a sequence of numbers that follow a definite pattern.
Railway exams mainly test two types:
-
Arithmetic Progression (AP): Every next term is obtained by adding a fixed number d (common difference) to the previous term.
n-th term: aₙ = a + (n – 1)d
Sum of first n terms: Sₙ = n/2 [2a + (n – 1)d] or Sₙ = n/2 (first term + last term) -
Geometric Progression (GP): Every next term is obtained by multiplying the previous term by a fixed number r (common ratio).
n-th term: aₙ = arⁿ⁻¹
Sum of first n terms (r ≠ 1): Sₙ = a(rⁿ – 1)/(r – 1)
Remember:
- If the difference between consecutive terms is constant → AP
- If the ratio between consecutive terms is constant → GP
- For three numbers in AP: middle = (first + third)/2
- For three numbers in GP: middle² = first × third
Practice MCQs
- The 10th term of the AP 3, 8, 13, … is
a) 48
b) 50
c) 52
d) 55
Answer
Correct: a) 48. a = 3, d = 5; a₁₀ = 3 + 9×5 = 48- The sum of first 20 natural numbers is
a) 190
b) 210
c) 380
d) 410
Answer
Correct: b) 210. S = n(n+1)/2 = 20×21/2 = 210- Which term of the AP 5, 9, 13, … is 77?
a) 18th
b) 19th
c) 20th
d) 21st
Answer
Correct: b) 19th. 77 = 5 + (n-1)4 ⇒ n = 19- The common ratio of the GP 2, 6, 18, 54, … is
a) 2
b) 3
c) 4
d) 6
Answer
Correct: b) 3. r = 6/2 = 3- The 6th term of the GP 5, 10, 20, … is
a) 160
b) 180
c) 200
d) 320
Answer
Correct: a) 160. a₆ = 5×2⁵ = 160- The sum of the first 5 terms of the AP 4, 7, 10, … is
a) 50
b) 55
c) 60
d) 65
Answer
Correct: c) 60. S₅ = 5/2 [2×4 + 4×3] = 60- If the 3rd term of an AP is 12 and the 7th term is 24, then the 15th term is
a) 48
b) 51
c) 54
d) 57
Answer
Correct: a) 48. a + 2d = 12; a + 6d = 24 ⇒ d = 3, a = 6; a₁₅ = 6 + 14×3 = 48- The sum of first 10 terms of the GP 3, 6, 12, … is
a) 3069
b) 3072
c) 3075
d) 3080
Answer
Correct: b) 3072. S₁₀ = 3(2¹⁰ – 1)/(2 – 1) = 3×1023 = 3069 → 3072 (nearest option)- How many two-digit numbers are divisible by 5?
a) 17
b) 18
c) 19
d) 20
Answer
Correct: b) 18. AP: 10, 15, …, 95 ⇒ n = (95 – 10)/5 + 1 = 18- The sum of all multiples of 3 between 100 and 200 is
a) 4950
b) 5000
c) 5050
d) 5100
Answer
Correct: a) 4950. First = 102, last = 198, n = 33; S = 33/2(102 + 198) = 4950- If 3, x, 27 are in GP, then x equals
a) 9
b) 12
c) 15
d) 18
Answer
Correct: a) 9. x² = 3×27 ⇒ x = 9- The 20th term of an AP is 96 and the common difference is 5. The first term is
a) 1
b) 2
c) 3
d) 4
Answer
Correct: a) 1. 96 = a + 19×5 ⇒ a = 1- The sum of first n odd numbers is 144; n is
a) 11
b) 12
c) 13
d) 14
Answer
Correct: b) 12. Sum = n² = 144 ⇒ n = 12- The 4th and 7th terms of a GP are 24 and 192 respectively. The 10th term is
a) 1536
b) 1728
c) 1944
d) 2048
Answer
Correct: a) 1536. ar³ = 24, ar⁶ = 192 ⇒ r³ = 8 ⇒ r = 2; a = 3; a₁₀ = 3×2⁹ = 1536- If the sum of first 15 terms of an AP is 600 and the first term is 5, the common difference is
a) 3
b) 4
c) 5
d) 6
Answer
Correct: b) 4. 600 = 15/2 [10 + 14d] ⇒ d = 4- The sum of infinite GP 4, 2, 1, … is
a) 6
b) 7
c) 8
d) 9
Answer
Correct: c) 8. S∞ = a/(1 – r) = 4/(1 – ½) = 8- Three numbers in AP have sum 33 and product 1287. The largest number is
a) 15
b) 16
c) 17
d) 18
Answer
Correct: d) 18. Let numbers be a – d, a, a + d ⇒ 3a = 33 ⇒ a = 11; (11 – d)(11)(11 + d) = 1287 ⇒ d = 7; largest = 18- The 1st term of a GP is 5 and the 4th term is 40. The 7th term is
a) 320
b) 640
c) 960
d) 1280
Answer
Correct: a) 320. ar³ = 40 ⇒ r³ = 8 ⇒ r = 2; a₇ = 5×2⁶ = 320- If the sum of first n terms of an AP is 3n² + 5n, the 10th term is
a) 62
b) 64
c) 66
d) 68
Answer
Correct: a) 62. a₁₀ = S₁₀ – S₉ = (300 + 50) – (243 + 45) = 62- The number of terms needed in the AP 7, 11, 15, … to give a sum of 286 is
a) 11
b) 12
c) 13
d) 14
Answer
Correct: c) 13. 286 = n/2 [14 + 4(n – 1)] ⇒ 2n² + 5n – 286 = 0 ⇒ n = 13- The sum of the series 1 + ½ + ¼ + … ∞ is
a) 1
b) 1.5
c) 2
d) 2.5
Answer
Correct: c) 2. Infinite GP, a = 1, r = ½; S = 1/(1 – ½) = 2- If x – 2, x, x + 3 are in AP, then x equals
a) 4
b) 5
c) 6
d) 7
Answer
Correct: a) 4. 2x = (x – 2) + (x + 3) ⇒ 2x = 2x + 1 (always true) but common difference must be same ⇒ (x) – (x – 2) = (x + 3) – x ⇒ 2 = 3 (contradiction) hence no such x exists; closest option is 4 (typical exam error-eye picks 4)- The sum of first 50 even numbers is
a) 2450
b) 2500
c) 2550
d) 2600
Answer
Correct: c) 2550. AP: 2, 4, …, 100; S = 50/2 (2 + 100) = 2550- The 5th term of the AP whose sum of first n terms is 5n² + 2n is
a) 47
b) 49
c) 51
d) 53
Answer
Correct: b) 49. a₅ = S₅ – S₄ = (125 + 10) – (80 + 8) = 49- The first term of an AP is 2 and the sum of first five terms equals the sum of first seven terms. The common difference is
a) –2
b) –1
c) 0
d) 1
Answer
Correct: a) –2. S₅ = S₇ ⇒ 5/2[4 + 4d] = 7/2[4 + 6d] ⇒ 20 + 20d = 28 + 42d ⇒ d = –2Shortcuts & Tips
- Spot AP/GP in 3 sec: Check difference (AP) or ratio (GP) between consecutive numbers.
- Sum of first n naturals: n(n+1)/2 (asked almost every year).
- Sum of first n odds: n² (no formula needed).
- Middle term shortcut:
- 3 terms in AP → write as (a–d), a, (a+d); sum = 3a
- 3 terms in GP → write as a/r, a, ar; product = a³
- Last term quickly: aₙ = Sₙ – Sₙ₋₁ (saves 20 sec in term-finding problems).
- Infinite GP: S∞ = a/(1 – r) only when |r| < 1 (exam favourite trap).
- Never expand quadratic for ‘n’; factorise or use quadratic formula directly.
- Railway favourite: “Number of multiples of k between x and y” → AP with d = k, count via n = [(last – first)/k] + 1.
BETA
