Application of A.P. and G.P. – Definitions, Examples, Applications

Application of A.P. and G.P.: An Arithmetic Progression (AP) is a set of terms in which the differences between each term are the same. Each successive term in a Geometric Progression (GP) is obtained by multiplying the common ratio by the preceding term. By using the basic knowledge about the application of A.P. and G.P, we can understand and apply these in our everyday life activities and find solutions to many problems in an easier way.

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Arithmetic Progression (A.P.)

A progression is a set of numbers that follow a set of rules. The difference between every two successive terms in an arithmetic progression (A.P.) is the same. It is possible to obtain a formula for the \({n^{{\rm{th}}}}\) term from an arithmetic progression.

An arithmetic progression is a set of terms in which the difference between each term is the same.

Each term, except the first, is obtained by adding a set integer to the previous term in an arithmetic progression.

Example: \(1, 5, 9, 13, 17, 21, 25, 29, 33,…\)

The sequence \(3, 7, 11, 15,…\), for example, is an arithmetic progression (A.P.) because each number is obtained by adding \(4\) to the previous term. The \({n^{{\text{th}}}}\) term in this series is \(4n-1\). By putting \(n=1,2,3,…\) in the \({n^{{\text{th}}}}\) term, you can get the sequence’s terms.

If \(n=1,4 n-1=3\)

If \(n=2,4 n-1=7\)

If \(n=3,4 n-1=11\)

If \(n=4,4 n-1=15\)

Learn About Arithmetic Progression

Arithmetic Progression (AP) Formula

Common difference of an arithmetic progression \(d=a_{2}-a_{1}\)

\({n^{{\text{th}}}}\) term of an arithmetic progression \(a_{n}=a+(n-1) d\)

Sum of \(n\) terms of an arithmetic progression \(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)

Geometric Progression (G.P.)

A geometric progression is a sequence in which each term has a fixed ratio, known as a common ratio. G.P. stands for geometric progression. The geometric sequence is generally written as \(a, a r, a r^{2} \ldots\), where \(a\) is the first term, and \(r\) is the sequence’s common ratio. Both negative and positive numbers are possible for the common ratio.

Each successive term in a geometric progression is obtained by multiplying the common ratio by the preceding term.

Example: \(6, 12, 24, 48,…\)

Geometric Progression (G.P) Formula

The common ratio of a geometric progression \(r=\frac{a_{2}}{a_{1}}\)

The formula for the \(n^{\text {th }}\) term of a geometric progression with the first term \(a\) and a common ratio of \(r\).

\(a_{n}=a r^{n-1}\)

The formula for the sum of geometric progression formula,

If \(|r|<1, S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}\)

If \(|r|>1, S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}\)

The sum of infinite geometric formula \(S_{\infty}=\frac{a}{1-r}\) where \(r<1\).

Applications of Arithmetic Progression

Some of the applications of an arithmetic progression of real life are:

1. A.P. is used in straight-line depreciation calculation
2. When someone is waiting for a cab, A.P. is employed to forecast any sequence. They can forecast when the next cab will arrive if the traffic is going at a consistent speed.
3. A.P. is used in Pyramid-like patterns, where things are constantly changing, and many more applications.
4. Checking accounts online is a fundamental application of daily arithmetic. With the risk of identity theft and online banking, general knowledge of fundamental mathematics is essential in such circumstances.
5. When you take a taxi, you can experience a real-life application of arithmetic progression. You will be charged an initial cost and then a per mile or kilometre charge once you have boarded a taxi. This diagram depicts an arithmetic sequence in which you will be charged a fixed (constant) rate plus the beginning rate for each kilometre travelled.

Applications of Geometric Progression

Some real-life applications of a geometric progression are listed below:

1. A finite geometric sequence is an example of a ball bouncing. The height of the ball is reduced by half each time it bounces. If the ball bounces from a height of \(4\) feet the next time, the highest bounce will be \(2\) feet, then \(1\), then \(6\) inches, and so on until the ball stops bouncing.
2. Geometric progression is commonly used in calculating the interest earned.
3. Geometric progression is commonly used in calculating the amount in our savings account.
4. Geometric progression helps to calculate the size of exponential population growth, such as bacteria in a Petri dish.
5. Mathematicians apply geometric series all of the time. They’re important in physics, engineering, biology, economics, computer science, queueing theory, and finance, among many other fields.

Examples of Application of A.P. and G.P.

Example of an Application of A.P.

An arithmetic progression is a set of terms in which the differences between each term are the same.

From the above example, we can write the arithmetic progression as

\(1, 3, 5, 7, 9, 11,…\)

Example of Application of G.P.:

A geometric sequence is a set of values in which each new term (save the first) is calculated by multiplying the previous term by a constant ratio \((r)\).

For example, assume one person was sick with the flu and failed to cover his mouth when two people visited them while he was in bed. They leave, and the sickness affects them the next day. Imagine that each friend transmits the virus to two friends the next day by the same droplet distribution. If this pattern continues and each sick person infects two more people, we can estimate the number of people infected.

From the above example, we can write the geometric progression as

\(1, 2, 4, 8, 16,…\)

Learn About Geometric Progression

Solved Examples – Application of A.P. and G.P.

Q.1. Find the \(6^{\text {th }}\) term of the G.P. \(\frac{1}{7}, \frac{1}{14}, \frac{1}{28}, \ldots\).
Ans: From the given \(a=\frac{1}{7}, r=\frac{\frac{1}{14}}{\frac{1}{7}}=\frac{1}{2}\)
We know the formula for the \({n^{{\rm{th}}}}\) term of G.P.
\(a_{n}=a r^{n-1}\)
\(\Rightarrow a_{6}=\frac{1}{7}\left(\frac{1}{2}\right)^{6-1}\)
\(\Rightarrow a_{6}=\frac{1}{7}\left(\frac{1}{2}\right)^{5}\)
\(\Rightarrow a_{6}=\frac{1}{7} \times \frac{1}{32}=\frac{1}{224}\)
Therefore, the \(6^{\text {th }}\) term of the given G.P is \(\frac{1}{224}\).

Q.2. Find the sum of the first \(12\) terms of the A.P.: \(8, 3,–2,…\)
Ans: Given AP: \(8,3,-2, \ldots\)
Here \(a=8\), common difference \(d=3-8=-5\) and \(n=12\)
We know,
Sum of \(n\) terms of an arithmetic progression \(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{12}=\frac{12}{2}[2(8)+(12-1)(-5)]\)
\(S_{12}=\frac{12}{2}[16+11(-5)]\)
\(S_{12}=6[16+11(-5)]\)
\(S_{12}=6[16-55]\)
\(S_{12}=6[-39]\)
\(S_{12}=-234\)
Therefore, the sum of the first \(12\) terms of the given A.P is \(-234\).

Q.3. The first term of an A.P. is \(3\), and the last term is \(17\). If the sum of all terms is \(150\), what is the \({4^{{\text{th}}}}\) term?
Ans: From the given \(a=3, l=17, S_{n}=150, a_{n}=\) ?
\(S_{n}=\frac{n}{2}[a+l]\)
\(\Rightarrow 150=\frac{n}{2}[3+17]\)
\(\Rightarrow 150=\frac{n}{2}[20]\)
\(\Rightarrow 150=10 n\)
\(\Rightarrow n=\frac{150}{10}\)
\(\Rightarrow n=15\)
Now, \(a_{15}=a+(n-1) d\)
\(\Longrightarrow 17=3+14 d\)
\(\Longrightarrow 14 d=14\)
\(\Rightarrow d=\frac{14}{14}=1\)
So, \(a_{4}=a+3 d\)
\(a_{4}=3+3(1)\)
\(a_{4}=3+3=6\)
Therefore, the \({4^{{\text{th}}}}\) term is \(6\).

Q.4. If the sum of the first \(14\) terms of an A.P. is \(1050\) and its first term is \(10\), find the \({15^{{\text{th}}}}\) term.
Ans: Given: \(S_{n}=1050, n=14\) and \(a=10\)
We know that the sum of \(n\) terms of an arithmetic progression \(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(\Rightarrow 1050=\frac{14}{2}[2(10)+(14-1) d]\)
\(\Rightarrow 1050=7[20+13 d]\)
\(\Rightarrow \frac{1050}{7}=20+13 d\)
\(\Rightarrow 150-20=13 d\)
\(\Rightarrow 13 d=130\)
\(\Rightarrow d=\frac{130}{13}\)
\(\Rightarrow d=10\)
So, \(15^{\text {th }}\) term of an arithmetic progression \(a_{15}=10+(14-1)(10)\)
\(\Rightarrow a_{15}=10+13(10)\)
\(\Rightarrow a_{15}=10+130\)
\(\Rightarrow a_{15}=140\)
Therefore, \(15^{\text {th }}\) term of an arithmetic progression is \(140\).

Q.5. Salary of Rohan, when his salary is \(₹ 5,00,000\)  for the first year and expected to receive a yearly increment of \(10%\). Now, find the Rohan salary at starting of the \({5^{{\text{th}}}}\) year.
Ans: From the given information, we have \(a=500000, r=1.1\)
Rohan salary at starting of \(5^{\text {th }}\) year is needed. So, \(n=5\)
We know the formula for the \(n^{\text {th }}\) term of G.P.
\(a_{n}=a r^{n-1}\)
\(\Rightarrow a_{n}=500000 \times(1.1)^{4}\)
\(\Rightarrow a_{n}=500000 \times 1.4641\)
\(\Rightarrow a_{n}=732050\)
Therefore, \(₹ 732050\) is the Rohan salary at starting of the \({5^{{\text{th}}}}\) year.

Summary 

Arithmetic progression can be applied in real life by understanding a specific pattern, including A.P. utilised in straight-line depreciation. Geometric sequences have many uses in everyday life, but one of the most common is calculating interest. This article includes the definition, formulas, examples, and applications of A.P. and G.P.

This article “Application of A.P. and G.P.” helps a lot in understanding the real-life applications of A.P. and G.P.

FAQs on Application of A.P. and G.P.

Q.1: What are the applications of arithmetic progression?
Ans: Some of the applications of an arithmetic progression are listed below.
1. A.P. used in straight-line depreciation calculation
2. A.P. is used when someone is waiting for a cab. A.P. is employed to forecast any sequence. They can forecast when the next cab will arrive if the traffic is going at a consistent speed.
3. A.P. is used in Pyramid-like patterns, where things are constantly changing, and many more applications.
4. A.P. is used in checking accounts online is a fundamental application of daily arithmetic.

Q.2: What is the application of geometric sequence in real life?
Ans: Geometric sequences have much application in everyday life, but calculating interest is one of the most prevalent uses. A term in a series is calculated by multiplying the sequence’s initial value by the rate raised to the power of one less than the term number.

Q.3: What are A.P. and G.P.? Give example.
Ans: An arithmetic progression is a set of terms in which the differences between each term are the same.
Each term, except the first, is obtained by adding a set integer to the previous term in an arithmetic progression.
Example: \(1, 3, 5, 7, 9,…\)
A geometric progression is a sequence in which each term has a fixed ratio, known as a common ratio. G.P. stands for geometric progression.
Example: \(4, 8, 16, 32, 64,…\)

Q.4: What is the formula of A.P. and G.P.?
Ans: The formulas of an arithmetic progression are,
\(n^{t h}\) term of an arithmetic progression \(a_{n}=a+(n-1) d\)
Sum of \(n\) terms of an arithmetic progression \(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
The formulas of a geometric progression are,
The formula for the nth term of a geometric progression \(a_{n}=a r^{n-1}\)
The formula for the sum of geometric progression formula. If \(\left| r \right| < 1,\,{S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}\)
If \(\left| r \right| > 1,\,{S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)

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