Arithmetic Progression and Geometric Progression

Arithmetic Progression and Geometric Progression: The word "sequence" in English means a collection of some numbers or objects in such a way that it has a first member, a second member, and so on. Sequences can be of anything, for example. - January, February, .... is the sequence of months in a year. Sequences come into use in actual real lives of people every day. Days of a week can also be considered as a sequence. Thus, it becomes essential to study the sequences and find patterns in them to predict the sequence's next terms and extract information from them. 

Sequences

Let's consider a sequence: 2,4,6,8 and so on. The various numbers occurring in it are called its terms. They are denoted by a₁, a₂, a₃ ... a_n.  The subscripts denote the nth term. The nth term of the sequence is also called the general term of the sequence because we can derive every other term from it by putting different values of n. Here in this case, 

a₁ = 2, a₂ = 4, a₃ = 6 and so on...

A sequence with a finite number of terms is called a finite sequence and similarly, a sequence with an infinite number of terms is called an infinite sequence. 

A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it. Sometimes, we use the functional notation a(n) for a_n. 

Series

For a given sequence a₁, a₂, a₃ ... a_n. The expression given below is called a series. A series can be infinite or finite depending upon the number of terms its sequence has. ∑ is the common notation used to denote the series. This indicates the summation involved. 

\Sigma^{n}_{i = 1}a_{i}  = a₁ + a₂+ a₃ +... a_n 

These concepts give rise to the sequences known as arithmetic progression and geometric progression. 

What is Arithmetic Progression (A.P)

Consider a sequence 1, 3, 5, 7, ..... Notice that in this sequence, the difference between successive terms is constant. This means that at each step a constant value is being added to each term of this sequence. A sequence a₁, a₂, a₃ ... a_n can be called an arithmetic progression if a_n+1 = a_n + d where n is any natural number. In such a series, a₁ is called the first term, and the constant term d is called the common difference of A.P. So, an AP looks like, 

a, a + d, a + 2d, a + 3d ..... and so on. 

The nth for AP can be defined as, 

a_n = a₁ + (n-1)d

Sum of n terms of an AP is given by, 

S_n = \frac{n}{2}[2a + (n-1)d]

or 

S_n  = \frac{n}{2}[a + l]

Read More: Arithmetic Progression

What is Geometric Progression (G.P)

Consider the following sequence, 2, 4, 8, 16 ..... It is clear here, that each term is being multiplied by 2 in this sequence. Such sequences where successive terms are multiplied by a constant number are called geometric progressions. In a more general way, a sequence a₁, a₂, a₃ ... a_n can be called a geometric progression if a_n+1 = a_n. r where n is any natural number. In such a series, a₁ is called the first term, and the constant term r is called the common ratio of G.P. So, a GP looks like, 

a, ar, ar², arⁿ..... and so on. 

The nth for GP can be defined as, 

a_n = a₁r^n-1

In general, GP can be finite and infinite but in the case of infinite GP, the common ratio must be between 0 and 1, or else the values of GP go up to infinity. Sum of GP consists of two cases: 

Let's denote the S_n are a + ar + ar² + ..... arⁿ

Case 1: If r = 1, the series collapses to 

a, a, a, a ... and so on. 

S_n = na

Case 2: If r≠1, the series stays the same, 

a + ar + ar² + ..... arⁿ

S_n = \frac{a(1 - r^n)}{1 - r}

Let's look at some word problems related to these concepts 

Arithmetic Progression vs Geometric Progression

Solved Examples on AP and GP

Question 1: A bitcoin stock started at $5. After that, every day it rises by $2. Find the stock price at the end of 16th day. 

Answer: 

In the above question, each time a constant number is added to the previous term to make a new term. This is an AP. 

5, 7, 9, ... and so on. 

Using the formula for nth term of AP. 

a_n = a₁ + (n-1)d

Here a₁ denotes the first term and d denotes the common difference. In this case ,

a₁ = 5, d = 2 and n = 16

a₁₀ = a₁ + (16-1)d
⇒ a₁₀ = 5 + (15)2
⇒ a₁₀ = 5 + 30
⇒ a₁₀ = 35 

Thus, the stock prices are at $35.

Question 2: A person planted 3 trees at his son's birth. After that, on subsequent birthdays he planted 5 more trees every year. Find the number of trees in his backyard when his son is 10years old. 

Answer: 

In the above question, each time a constant number is added to the previous term to make a new term. This is an AP. 

3, 8, 13, . . . and so on. 

Using the formula for nth term of AP. 

a_n = a₁ + (n-1)d

Here a₁ denotes the first term and d denotes the common difference. In this case,

a₁ = 3, d = 5 and n = 10

As 3 trees planted at birth, and 5 on each birthday, he plants 10 times,

a₁₁ = a₁ + (11-1)d
⇒ a₁₁ = a₁ + (10)d
⇒ a₁₁ = 3 + 10(5)
⇒ a₁₁ = 3 + 50 
⇒ a₁₁ = 53

Thus, there are 53 trees in his backyard now. 

Question 3: English rock band the1975  released a new album in summer, and they opened with 100,000 copies sold within one day. Now the album is topping charts and everyday they sell 20,000 copies more than the previous day. Find the total album sales in a week. 

Answer: 

In the above question, each time a constant number is added to the previous term to make a new term. This is an AP. 

100,000; 120,000; 140,000; ... and so on. 

Goal is to calculate the sum of the sequence at the end of 10th day. 

Using the formula for sum till nth term of AP. 

S_n = \frac{n}{2}[2a + (n-1)d]

Here a denotes the first term and d denotes the common difference. In this case,

a = 100,000, d = 20,000 and n = 7

S_n = \frac{n}{2}[2a + (n-1)d]
⇒ S₇ = \frac{7}{2}[2*100000 + (7-1)(20000)]
⇒ S₇ = \frac{7}{2}[200000 + (6)(20000)]
⇒ S₇ = \frac{7}{2}[200000 + (120000)]
⇒ S₇ = \frac{7}{2}[320000]
⇒ S₇ = 7(160000)
⇒ S₇ = 1,120,000

Thus, the total album sale is 1,120,000.

Question 4: The deer population is increasing in Corbett National Park. In year 2015 it was 1000, since then it has been increasing, and it becomes 2 times every year. Find the population in 2021. 

Solution. 

 Here, every year the population becomes 2 times. A constant number is being multiplied to the previous term to get the new term. This is a geometric progression. 

1000, 2000 ... and so on. 

Here a = 1000 and r= 2

Using the formula for nth term of the GP 

a_n = a₁r^n-1

In 2021, n = 7. Plugging in the values in the formula 

a_n = a₁r^n-1
⇒ a_n = (1000)(2)^(7-1)
⇒ a_n = (1000)(2)⁶
⇒ a_n = (1000)(64) 
⇒ a_n = 64000 

There must be 64,000 deer in Corbett National Park now. 

Question 5: A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the number of ancestors in last 10 generations of this family.  

Solution. 

 Here, every year the number becomes 2 times. A constant number is being multiplied to the previous term to get the new term. This is a geometric progression. 

2,4 ... and so on. 

Here a = 2 and r= 2

Using the formula for nth term of the GP 

a_n = a₁r^n-1

In 2021, n = 10. Plugging in the values in the formula 

a_n = a₁r^n-1
⇒ a_n = (2)(2)^(10-1)
⇒ a_n = (2)(2)⁹
⇒ a_n = (2)¹⁰
⇒ a_n = 1024

The total number of ancestors in the last 10 generations is . 

Question 6: Insert two numbers between 4 and 256 such that the resulting sequence becomes a GP. 

Answer: 

Let's say those two numbers are x and y. The resulting sequence then becomes, 

4, x, y, 256 

This sequence has four terms and is a GP. Here, 

a = 4 and r = ? 

Formula for nth term of GP is  

a_n = a₁r^n-1

4th term is 256, 
⇒ 256 = 4r^{(4 - 1)}
⇒ 64 = r³

This means that, r = 4

Thus, 

x = ar 
⇒ x = (4)(4) 
⇒ x= 16

and y = ar²
⇒ y = 4(4)²
⇒ y = 64

So, the two numbers to be inserted are 16 and 64 

Question 7: Number of bacteria in a dish are 100, and they are increasing by double the previous value every hour. Find the number of bacteria in the dish after 6 hours. 

Answer: 

 Here, every year the number becomes 2 times. A constant number is being multiplied to the previous term to get the new term. This is a geometric progression. 

100,200, 400 ... and so on. 

Here a = 100 and r= 2

Using the formula for sum till nth term of the GP 

S_n = \frac{a(1 - r^n)}{1 - r}

n = 6. Plugging in the values in the formula 

S_n = \frac{a(1 - r^n)}{1 - r}
⇒ S_n = \frac{a(1 - r^n)}{1 - r}
⇒ S₆ = \frac{100(1 - 2^6)}{1 - 2}
⇒ S₆ = \frac{100(2^6 - 1)}{2 - 1}
⇒ S₆ = \frac{100(64 - 1)}{2 - 1}
⇒ S₆ = 6300

There must be 63,00 bacteria in the dish now. 

Practice Questions on AP and GP

Question 1. Find the 10th term of the arithmetic progression (AP) where the first term a = 3 and the common difference d = 5.

Question 2. Find the sum of the first 20 terms of the arithmetic progression where the first term a = 2 and the common difference d = 3.

Question 3. Given the arithmetic progression 7, 13, 19,…, find the common difference.

Question 4. Find the 5th term of the geometric progression (GP) where the first term a = 2 and the common ratio r = 3.