Arithmetic and Geometric Progressions
Arithmetic and Geometric Progressions
An arithmetic progression (AP) adds the same number each time; a geometric progression (GP) multiplies by the same number. NTPC asks the nth term, which term equals a given value, sums of terms, the sum of multiples in a range and, in CBT-2, means and infinite GP sums.
1. Formula Box
| Arithmetic progression | Geometric progression | |
|---|---|---|
| Form | a, a + d, a + 2d, ... | a, ar, ar², ... |
| nth term | a + (n − 1)d | ar^(n − 1) |
| Sum of n terms | (n/2)[2a + (n − 1)d] = (n/2)(first + last) | a(rⁿ − 1)/(r − 1), r ≠ 1 |
| Infinite sum | Not defined (unless all zero) | a/(1 − r) when −1 < r < 1 |
| Mean of p and q | AM = (p + q)/2 | GM = √(pq) |
Two terms of an AP: d = (Tₘ − Tₙ)/(m − n) Two terms of a GP: r^(m − n) = Tₘ/Tₙ
AM ≥ GM for positive numbers, equal only when the numbers are equal
Sum of first n natural numbers = n(n + 1)/2; of first n odd numbers = n²
2. Shortcut Rules
| Situation | Rule |
|---|---|
| Multiples of k between two numbers | First and last multiple, then n = (last − first)/k + 1 |
| Three numbers in AP | Take them as a − d, a, a + d |
| Three numbers in GP | Take them as a/r, a, ar |
| Quadratic in n for a sum | Both roots may be valid if a later term is zero |
✗ Number of terms in 7, 11, ..., 83 = (83 − 7)/4 | ✓ Add 1: (83 − 7)/4 + 1 = 20
✗ 8th term of 3, 6, 12, ... = 3 × 2⁸ | ✓ The power is n − 1: 3 × 2⁷ = 384
हिंदी नोट: समांतर श्रेढ़ी का n-वाँ पद a + (n − 1)d और योग (n/2)(पहला + अंतिम) होता है। गुणोत्तर श्रेढ़ी का n-वाँ पद arⁿ⁻¹ और अनंत योग a/(1 − r), जब r का मान −1 और 1 के बीच हो।
Exam Pointer: Verified NTPC pattern: the nth term of a geometric series from its first few terms (June 2022 CBT-2). AP term and sum questions and CBT-2 mean and infinite GP questions had no verified NTPC shift in our check and are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: nth term of an AP and which term equals a value
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the 20th term of the AP 7, 11, 15, ...
a = 7 and d = 4, so T₂₀ = 7 + 19 × 4 = 83.
Answer: 83
EXAMATLAS LEVEL
Q. The 5th term of an AP is 23 and the 12th term is 51. Find the 30th term, and which term equals 111.
d = (51 − 23)/(12 − 5) = 4 and a = 23 − 4 × 4 = 7. T₃₀ = 7 + 29 × 4 = 123. 7 + (n − 1) × 4 = 111 gives n − 1 = 26, so n = 27.
Answer: 123; the 27th term
Pattern 2: Sum of an AP and sums of multiples
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the sum of all numbers between 100 and 200 that are divisible by 7.
The first is 105 and the last 196, so n = (196 − 105)/7 + 1 = 14. Sum = (14/2)(105 + 196) = 7 × 301 = 2,107.
Answer: 2,107
EXAMATLAS LEVEL
Q. How many terms of the AP 54, 51, 48, ... must be taken for the sum to be 513?
(n/2)[108 − 3(n − 1)] = 513 gives n(111 − 3n) = 1,026, that is n² − 37n + 342 = 0 = (n − 18)(n − 19). Both work: the 19th term is 54 − 18 × 3 = 0, so adding it leaves the sum unchanged.
Answer: 18 or 19 terms
Pattern 3: nth term and sum of a GP
[PYQ: NTPC CBT-2 14-Jun-2022 Shift-1]
EXAM LEVEL
Q. Find the 8th term and the sum of the first 8 terms of the GP 3, 6, 12, ...
a = 3 and r = 2. T₈ = 3 × 2⁷ = 384. S₈ = 3(2⁸ − 1)/(2 − 1) = 3 × 255 = 765.
Answer: 384 and 765
EXAMATLAS LEVEL
Q. The 3rd term of a GP is 12 and the 6th term is 96. Find the first term and the sum of the first 7 terms.
r³ = 96/12 = 8, so r = 2 and a = 12/4 = 3. S₇ = 3(2⁷ − 1) = 3 × 127 = 381.
Answer: 3 and 381
CBT-2 LEVEL
Pattern 4: Means and the infinite GP
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the AM and GM of 4 and 16, and check AM ≥ GM.
AM = 20/2 = 10 and GM = √64 = 8. Indeed 10 ≥ 8.
Answer: AM 10, GM 8
EXAMATLAS LEVEL
Q. An infinite GP has first term 5 and sum 20. Find its common ratio and the sum of the squares of all its terms.
5/(1 − r) = 20 gives 1 − r = 1/4, so r = 3/4. The squares form a GP with first term 25 and ratio 9/16, so their sum = 25/(1 − 9/16) = 25 × 16/7 = 400/7.
Answer: r = 3/4; 400/7
60-Second Revision
- AP: Tₙ = a + (n − 1)d; Sₙ = (n/2)(first + last).
- Count terms as (last − first)/d + 1.
- GP: Tₙ = arⁿ⁻¹; Sₙ = a(rⁿ − 1)/(r − 1).
- Infinite GP: a/(1 − r) for −1 < r < 1.
- AM ≥ GM for positive numbers.