Mathematics Classes VI-X for School Assistants and TET Paper 2A · Chapter 5

Arithmetic and Geometric Progressions

What to remember

In an arithmetic progression (AP) each term is got by adding a fixed number d, the common difference. The nth term is a + (n − 1)d and the sum of n terms is (n/2)[2a + (n − 1)d].
In a geometric progression (GP) each term is got by multiplying by a fixed number r, the common ratio. The nth term is ar^(n−1) and the sum of n terms is a(rⁿ − 1)/(r − 1) when r ≠ 1.
Three terms a, b, c are in AP if 2b = a + c, and in GP if b² = ac. These two tests solve many problems quickly.

Sequences and series

A sequence is a list of numbers in a definite order, such as 2, 4, 6, 8, …. Each number is a term. A series is the sum of the terms. The first term is a₁ or a, and the nth term is aₙ. Many sequences follow a rule; AP and GP are the two most important rules.

Arithmetic progression

A list of numbers in which the difference between each term and the one before it is always the same. That difference is the common difference d = a₂ − a₁ = a₃ − a₂ and so on. If d > 0 the AP increases, if d < 0 it decreases, and if d = 0 all terms are equal.

Examples: 3, 7, 11, 15, … (a = 3, d = 4); 20, 17, 14, … (a = 20, d = −3); 5, 5, 5, … (d = 0).

Key formulas for an AP with first term a and common difference d:

Quantity	Formula
nth term	aₙ = a + (n − 1)d
nth term from the end	l − (n − 1)d, where l is the last term
Number of terms	n = (l − a)/d + 1
Sum of first n terms	Sₙ = (n/2)[2a + (n − 1)d]
Sum using last term	Sₙ = (n/2)(a + l)
Term from sum	aₙ = Sₙ − Sₙ₋₁
Test for three terms	a, b, c in AP if 2b = a + c

Worked example 1: How many terms are in 7, 13, 19, …, 205? d = 6 and n = (205 − 7)/6 + 1 = 34.

Worked example 2: The sum of the first 20 terms of 3, 7, 11, … is S₂₀ = 10[6 + 19 × 4] = 10 × 82 = 820.

Worked example 3: The 10th term of an AP is 52 and the 17th term is 94. Then a + 9d = 52 and a + 16d = 94; subtracting gives 7d = 42, so d = 6 and a = −2.

Worked example 4: The sum of the first n natural numbers is n(n + 1)/2. For n = 100 this is 5050. A well-known classroom story, the pairing of 1 + 100, 2 + 99, …, shows why: there are 50 pairs, each with sum 101.

Choosing terms: for three numbers in AP use a − d, a, a + d (their sum is 3a). For four numbers use a − 3d, a − d, a + d, a + 3d.

Arithmetic mean and special sums

The arithmetic mean of a and b is (a + b)/2. It is the middle term of the AP a, (a + b)/2, b. To insert n arithmetic means between a and b, the common difference is d = (b − a)/(n + 1). Example: three means between 2 and 18 give d = 16/4 = 4, so the terms are 2, 6, 10, 14, 18.

Sum	Formula
1 + 2 + … + n	n(n + 1)/2
1 + 3 + 5 + … (n odd numbers)	n²
2 + 4 + 6 + … (n even numbers)	n(n + 1)
1² + 2² + … + n²	n(n + 1)(2n + 1)/6
1³ + 2³ + … + n³	[n(n + 1)/2]²

Geometric progression

A list of non-zero numbers in which each term after the first is the previous term multiplied by a fixed number r, the common ratio. So r = a₂/a₁ = a₃/a₂ and so on.

Examples: 2, 6, 18, 54, … (a = 2, r = 3); 80, 40, 20, 10, … (a = 80, r = 1/2); 3, −6, 12, −24, … (r = −2, alternating signs).

Quantity	Formula
nth term	aₙ = a rⁿ⁻¹
Sum of n terms (r > 1)	Sₙ = a(rⁿ − 1)/(r − 1)
Sum of n terms (r < 1)	Sₙ = a(1 − rⁿ)/(1 − r)
Sum when r = 1	Sₙ = na
Sum to infinity (	r	< 1)	S∞ = a/(1 − r)
Test for three terms	a, b, c in GP if b² = ac
Geometric mean of a and b	√(ab)

Worked example 5: In the GP 2, 6, 18, … the 5th term is 2 × 3⁴ = 162, and the sum of the first 5 terms is 2(3⁵ − 1)/(3 − 1) = 242.

Worked example 6: Find the sum of 1 + 1/2 + 1/4 + …. Here a = 1 and r = 1/2, so S∞ = 1/(1 − 1/2) = 2.

Worked example 7: The 3rd term of a GP is 12 and the 6th term is 96. Then ar² = 12 and ar⁵ = 96; dividing gives r³ = 8, so r = 2 and a = 3.

Worked example 8: A ball loses half of its height after each bounce. A teacher can use this to introduce a GP with r = 1/2 and to show that infinite sums can be finite.

Choosing terms: for three numbers in GP use a/r, a, ar (their product is a³). To insert n geometric means between a and b, the common ratio is (b/a)^(1/(n + 1)).

AP and GP compared

Feature	AP	GP
Rule	Add d each time	Multiply by r each time
General term	a + (n − 1)d	ar^(n−1)
Mean of two terms	AM = (a + b)/2	GM = √(ab)
Test for b being the middle term	2b = a + c	b² = ac
Graph of terms against n	Points on a straight line	Points on an exponential curve
Sum to infinity	Does not exist (unless all terms are 0)	a/(1 − r) when	r	< 1

For two positive numbers, the AM is at least as large as the GM: (a + b)/2 ≥ √(ab), with equality only when a = b.

Classroom angle

Let students build the patterns with objects: matchsticks for an AP and paper folding for a GP (the number of layers doubles with each fold). Ask them to guess the next term first and then state the rule. Common errors: using n instead of n − 1 in the nth term formula, and finding the "ratio" by subtraction. Students should check d by finding the difference between several consecutive pairs.

Exam traps

n versus n − 1: aₙ = a + (n − 1)d, not a + nd.
Common ratio: r is found by division, d by subtraction.
Sum formula choice: use (n/2)(a + l) when the last term is known, and (n/2)[2a + (n − 1)d] when d is known.
Infinite sum: exists only for |r| < 1; for r = 2 it does not exist.
Zero common difference: a constant list is an AP with d = 0.
Sign of r: a negative r gives alternating signs in the GP.
Sum of n odd numbers: it is n², not n(n + 1).
Test for GP: b² = ac; do not confuse it with 2b = a + c.

One-liners

1. In an AP the difference between consecutive terms is constant.
2. The nth term of an AP is a + (n − 1)d.
3. The sum of the first n terms of an AP is (n/2)[2a + (n − 1)d].
4. Sum of the first n natural numbers = n(n + 1)/2.
5. Sum of the first n odd numbers = n².
6. a, b, c are in AP when 2b = a + c.
7. In a GP the ratio of consecutive terms is constant.
8. The nth term of a GP is ar^(n−1).
9. The sum of n terms of a GP (r ≠ 1) is a(rⁿ − 1)/(r − 1).
10. S∞ = a/(1 − r) for |r| < 1.
11. a, b, c are in GP when b² = ac.
12. For positive numbers, AM ≥ GM.

Practice questions

The nth term of an AP with first term a and common difference d is
1. a × dⁿ⁻¹
2. a + (n − 1)d
3. a + nd
4. a + (n + 1)d
Answer
B. a + (n − 1)d
Standard formula.
The common difference of the AP 3, 7, 11, 15, … is
1. 3
2. 4
3. −4
4. 7
Answer
B. 4
7 − 3 = 4.
The 10th term of the AP 2, 5, 8, … is
1. 26
2. 32
3. 27
4. 29
Answer
D. 29
2 + 9 × 3 = 29.
The number of terms in the AP 7, 13, 19, …, 205 is
1. 33
2. 35
3. 34
4. 30
Answer
C. 34
n = (205 − 7)/6 + 1 = 34.
The sum of the first 20 terms of the AP 3, 7, 11, … is
1. 860
2. 820
3. 410
4. 800
Answer
B. 820
S₂₀ = 10[6 + 19 × 4] = 820.
The sum of the first 100 natural numbers is
1. 4950
2. 10100
3. 5000
4. 5050
Answer
D. 5050
n(n + 1)/2 = 100 × 101/2.
The sum of the first 10 odd natural numbers is
1. 55
2. 100
3. 90
4. 110
Answer
B. 100
The sum of n odd numbers is n² = 100.
The sum of the first 10 even natural numbers is
1. 110
2. 55
3. 90
4. 100
Answer
A. 110
n(n + 1) = 10 × 11.
The common ratio of the GP 2, 6, 18, 54, … is
1. 3
2. 4
3. 2
4. 9
Answer
A. 3
6/2 = 3.
The 5th term of the GP 2, 6, 18, … is
1. 486
2. 81
3. 54
4. 162
Answer
D. 162
2 × 3⁴ = 162.
The sum of the first 5 terms of the GP 2, 6, 18, … is
1. 484
2. 121
3. 242
4. 162
Answer
C. 242
2(3⁵ − 1)/(3 − 1) = 242.
The sum of the infinite series 1 + 1/2 + 1/4 + … is
1. 1
2. 3/2
3. Infinite
4. 2
Answer
D. 2
a/(1 − r) = 1/(1/2) = 2.
The 3rd term of a GP is 12 and its 6th term is 96. The first term is
1. 3
2. 6
3. 2
4. 4
Answer
A. 3
r³ = 8, r = 2, and a × 4 = 12 gives a = 3.
The value of x for which 2x, x + 10 and 3x + 2 are in AP is
1. 4
2. 6
3. 8
4. 10
Answer
B. 6
2(x + 10) = 2x + 3x + 2 gives 3x = 18.
If 4, x, 36 are in GP with x positive, then x is
1. 20
2. 18
3. 12
4. 16
Answer
C. 12
x² = 4 × 36 = 144.
The arithmetic mean of 6 and 14 is
1. 8.4
2. 10
3. 20
4. 9
Answer
B. 10
(6 + 14)/2 = 10.
The geometric mean of 4 and 9 is
1. 36
2. 6.5
3. 13
4. 6
Answer
D. 6
√(4 × 9) = 6.
When three arithmetic means are inserted between 2 and 18, the common difference is
1. 5
2. 4
3. 16/3
4. 3
Answer
B. 4
d = (18 − 2)/4 = 4.
The 10th term of an AP is 52 and the 17th term is 94. The common difference is
1. 5
2. 8
3. 6
4. 7
Answer
C. 6
7d = 42, so d = 6.
The 30th term of the AP 10, 7, 4, … is
1. 77
2. −80
3. −87
4. −77
Answer
D. −77
10 + 29 × (−3) = −77.
Which of the following lists is an AP?
1. 2, 4, 8, 16, …
2. 1, 1/2, 1/3, …
3. 1, 4, 7, 10, …
4. 1, 4, 9, 16, …
Answer
C. 1, 4, 7, 10, …
The difference is 3 each time.
Which of the following lists is a GP?
1. 3, 6, 12, 24, …
2. 1, 4, 9, 16, …
3. 3, 6, 9, 12, …
4. 2, 3, 5, 8, …
Answer
A. 3, 6, 12, 24, …
The ratio is 2 each time.
The value of 1² + 2² + … + 10² is
1. 55
2. 440
3. 385
4. 3025
Answer
C. 385
n(n + 1)(2n + 1)/6 = 10 × 11 × 21/6 = 385.
The value of 1³ + 2³ + 3³ + 4³ + 5³ is
1. 225
2. 125
3. 55
4. 15
Answer
A. 225
[5 × 6/2]² = 15² = 225.
Three numbers in AP have sum 15 and product 105. The smallest of them is
1. 3
2. 1
3. 5
4. 7
Answer
A. 3
The numbers are 3, 5, 7.
If the sum of n terms of an AP is n² + 2n, the 10th term is
1. 23
2. 21
3. 120
4. 20
Answer
B. 21
S₁₀ − S₉ = 120 − 99 = 21.
The sum to infinity of the GP 4, −2, 1, … is
1. 8
2. 4/3
3. 8/3
4. 2
Answer
C. 8/3
a/(1 − r) = 4/(3/2) = 8/3.
The sum of the first 8 terms of the GP 1, 2, 4, … is
1. 511
2. 256
3. 127
4. 255
Answer
D. 255
(2⁸ − 1)/(2 − 1) = 255.
The 6th term of the GP 80, 40, 20, … is
1. 5/2
2. 5
3. 10
4. 2
Answer
A. 5/2
80 × (1/2)⁵ = 2.5.
The number of two-digit numbers divisible by 4 is
1. 23
2. 22
3. 21
4. 20
Answer
B. 22
The numbers are 12 to 96: (96 − 12)/4 + 1 = 22.
The sum of the first 15 multiples of 3 is
1. 345
2. 720
3. 180
4. 360
Answer
D. 360
3 × 15 × 16/2 = 360.
The sum of 5 terms of an AP whose third term is 8 is
1. 24
2. 40
3. 32
4. 8
Answer
B. 40
S₅ = 5 × (middle term) = 40.
The sum 5 + 9 + 13 + … + 41 is
1. 236
2. 205
3. 46
4. 230
Answer
D. 230
n = 10 and S = 10/2 × (5 + 41) = 230.
The AM of 4 and 16 exceeds their GM by
1. 1
2. 2
3. 4
4. 6
Answer
B. 2
AM = 10 and GM = 8.
The sum of the series 1 − 1/3 + 1/9 − … to infinity is
1. 3/2
2. 2/3
3. 3/4
4. 1
Answer
C. 3/4
a = 1, r = −1/3, so S = 1/(4/3) = 3/4.
The third term from the end of the AP 5, 8, 11, …, 95 is
1. 89
2. 92
3. 83
4. 86
Answer
A. 89
95 − 2 × 3 = 89.
Consider the statements: 1. In an AP the difference of consecutive terms is constant. 2. In a GP the common ratio is found by subtracting consecutive terms.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
In a GP the ratio is found by division.
Consider the statements: 1. a, b, c are in AP if 2b = a + c. 2. a, b, c are in GP if b² = ac.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are the standard tests.
Consider the statements: 1. The sum to infinity of the GP 1, 2, 4, … exists. 2. The sum to infinity of a GP exists when |r| < 1.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
With r = 2 the terms grow without limit.
Consider the statements: 1. The sum of the first n odd numbers is n(n + 1). 2. The sum of the first n even numbers is n(n + 1).
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
The sum of n odd numbers is n².
Consider the statements: 1. A list of equal numbers such as 5, 5, 5, … is an AP with d = 0. 2. A GP can have a negative common ratio.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are true.
Consider the statements: 1. aₙ = a + (n − 1)d for an AP. 2. aₙ = a rⁿ for a GP.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
The GP formula is a r^(n−1).
Consider the statements: 1. The common difference of an AP is found by dividing consecutive terms. 2. S∞ = a/(1 − r) holds even when r = 1.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
D. Neither 1 nor 2
d is found by subtracting; r = 1 does not satisfy |r| < 1.
Consider the statements: 1. For positive numbers a and b, AM ≥ GM. 2. AM = GM only when a = b.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Equality holds only for equal numbers.
Consider the statements: 1. The common ratio of 2, −4, 8, … is −2. 2. The terms of this GP all have the same sign.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
The signs alternate when r is negative.

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1.The ratio of milk and water in a 30 litre mixture is 3: 2. Find the quantity of water to be added to the mixture in order to make this ratio 1:1.
2.There are 2822 students in a school and the ratio of boys to girls in the school is 36: 47, then find the number of boys in school.
3.There are 2664 students in a school and the ratio of boys to girls in the school is 41: 31, then find the number of boys in school.
4.Three partners invested in a business in the ratio 6:1:5. They invested their capitals for 7 months, 10 months and 8 months, respectively. What was the ratio of their profits? B y
5.The least common multiple of two numbers is 364 and their greatest common factor is 26. If one of the numbers is 26, then find the other number.
6.The monthly incomes of two friends Keshav and Tushar, are in the ratio 5: 7 respectively and each of them saves ₹93000 every month. If the ratio of their monthly expenditure is 2: 3, find the monthly income of Keshav(in ₹).
7.The monthly incomes of two friends Kiran and Mahesh, are in the ratio 5: 7 respectively and each of them saves ₹75000 every month. If the ratio of their monthly expenditure is 2: 4, find the monthly income of Kiran(in ₹).
8.The monthly incomes of two friends Ashish and Kiran, are in the ratio 5: 7 respectively and each of them saves ₹81000 every month. If the ratio of their monthly expenditure is 2: 3, find the monthly income of Ashish(in ₹).
9.The monthly incomes of two friends Nirmal and Rakesh, are in the ratio 5: 7 respectively and each of them saves ₹84000 every month. If the ratio of their monthly expenditure is 2: 4, find the monthly income of Nirmal(in ₹).
10.The monthly incomes of two friends Chetan and Vipul, are in the ratio 5: 7 respectively and each of them saves ₹72000 every month. If the ratio of their monthly expenditure is 1: 3, find the monthly income of Chetan(in ₹).

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