Geometric Progression (GP)

A geometric progression, also known as a geometric sequence, is a mathematical sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with a common ratio of 3. In real life, we use geometric progressions to calculate the size of exponential population growth, such as bacteria in a container.

This Story also Contains

1. What is Geometric Progression?
2. Increasing and Decreasing GP
3. Properties of Geometric Progression
4. Solved Examples Based on Geometric Progression

In this article, we will cover the concept of the Geometric Progression. This category falls under the broader category of Sequence and series, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of 16 questions have been asked on this concept, including one in 2013, one in 2017, two in 2019, three in 2020, four in 2021, two in 2022 and four in 2023

What is Geometric Progression?

A geometric progression or geometric sequence is a sequence where the first term is non-zero and the ratio between consecutive terms is always constant. The ‘constant factor’ is called the common ratio and is denoted by ‘r’. r is also a non-zero number.

The first term of a G.P. is usually denoted by 'a'.

If $a_1, a_2, a_3 \ldots . a_{n-1}, a_n$ is in geometric progression then, $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}=\ldots . .=\frac{a_n}{a_{n-1}}$

Eg,
- $2,6,18,54, \ldots .(a=2, r=3)$
- $4,2,1,1 / 2,1 / 4, \ldots .(a=4, r=1 / 2)$
- $-5,5,-5,5, \ldots \ldots .(a=-5, r=-1)$

General Term of a GP

If ' $a$ ' is the first term and ' $r$ ' is the common ratio, then
$
\begin{aligned}
& a_1=a=a r^{1-1}\left(1^{\text {st }} \text { term }\right) \\
& a_2=a r=a r^{2-1}\left(2^{\text {nd }} \text { term }\right) \\
& a_3=a r^2=a r^{3-1}\left(3^{\text {rd }} \text { term }\right) \\
& \cdots \\
& \cdots \\
& a_n=a r^{n-1}\left(\mathrm{n}^{\text {th }} \text { term }\right)
\end{aligned}
$

So, the general term or $\mathrm{n}^{\text {th }}$ term of a geometric progression is $a_n=a r^{n-1}$

Increasing and Decreasing GP

For a GP to be increasing or decreasing, r > 0. Since, If r < 0, then the terms of G.P. are alternately If a > 0, then G.P. is increasing if r > 1 and decreasing if 0 <r<1.

If a < 0 then G.P. is decreasing If r > and increasing if 0 < r <1positive and negative so neither increasing nor decreasing.

Properties of Geometric Progression

1. If a, b, c are in GP, then b² = a.c

2. If each term of a G.P. is multiplied by a fixed constant or divided by a non-zero fixed constant then the resulting series is also in G.P. with the same common ratio as the original series.

3. If each term of a G.P. is raised to some real number m, then the resulting series is also in G.P.

4. If two series are in GP then the the product of the series is also in GP.

If $a_1, a_2, a_3 \ldots$ and $b_1, b_2, b_3 \ldots$...are two G.P.'s, then $a_1 b_1, a_2 b_2, a_3 b_3 \ldots \ldots$ and $\frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3} \ldots$ are also G.P.
5. If $a_1, a_2, a_3, \ldots ., a_{n-1}, a_n$ are in G.P. with common ratio $r$, then is in A.P. and the converse also holds true.
6. If three numbers in G.P. whose product is given are to be taken, then take them as $\mathrm{a} / \mathrm{r}, \mathrm{a}$, ar.
7. If four numbers in G.P. whose product is given are to be taken, then take them as $\frac{a}{r^3}, \frac{a}{r}, a r, a r^3$.

8. The product of terms equidistant from the start and end of the G.P. is constant and it equals the product of the first and the last terms.

Recommended Video Based on Geometric Progression

Solved Examples Based on Geometric Progression

Example 1: Let $0<z<y<x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, \sqrt{2} y, z$ are in a geometric progression. If $x y+y z+z x=\frac{3}{\sqrt{2}} x y z$, then $3(x+y+z)^2$ [JEEMAINS 2023]
Solution:
$
\begin{aligned}
& 2 y^2=x z \\
& \frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2} \quad 4 y^2+2 y^2=\frac{3}{\sqrt{2}} y \cdot 2 y^2 \\
& x+z=4 y \quad 6 y^2=3 \sqrt{2} y^3 \\
& x y+y z+z x=\frac{3}{\sqrt{2}} x y z \quad \begin{array}{l}
y=\sqrt{2} \\
x+y+z=5 y=5 \sqrt{2}
\end{array} \\
& y(x+z)+z x=\frac{3}{\sqrt{2}} x z \cdot y^{3(x+y+z)^2}=3 \times 50=150 \\
&
\end{aligned}
$

Hence, the required answer is 150.

Example 2: For the two positive numbers $\mathrm{a}, \mathrm{b}$ is $\mathrm{a}$, and $\mathrm{b}$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{\mathrm{a}}, 10$ and $\frac{1}{\mathrm{~b}}$ are in an arithmetic progression, then $16 \mathrm{a}+\mathrm{b}$ is equal to [JEE MAINS 2023]
Solution:
$
\begin{aligned}
& \mathrm{b}^2=\frac{a}{18} \\
& 20=\frac{1}{a}+\frac{1}{b} \\
& a=\frac{b}{20 b-1} \\
& b^2=\frac{1}{18} \times \frac{b}{20 b-1} \\
& 360 b^2-18 b-1=0 \\
& 360 b^2-30 b+12 b-1=0 \\
& (12 b-1)(30 \mathrm{~b}+1)=0 \\
& b=\frac{1}{12}, \frac{-1}{30}(\text { rejected }) \\
& a=\frac{1}{8} \\
& 16 \mathrm{a}+12 \mathrm{~b}=2+1=3
\end{aligned}
$

Hence, the required answer is 3.

Example 3: Let the first term a and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three is 33033, then the sum of these terms is equal to : [JEE MAINS 2023]

Solution:

Let $a, a r, a r^2$ be three terms of GP
Given : $\mathrm{a}^2+(a r)^2+\left(\mathrm{ar}^2\right)^2=33033$
$
\begin{aligned}
& a^2\left(1+r^2+r^4\right)=11^2 \cdot 3.7 .13 \\
& \Rightarrow \mathrm{a}=11 \text { and } 1+r^2+r^4=3.7 .13 \\
& \Rightarrow r^2\left(1+r^2\right)=273-1 \\
& \Rightarrow r^2\left(r^2+1\right)=272=16 \times 17 \\
& \Rightarrow r^2=16 \\
& \therefore r=4 \quad[\because r>0]
\end{aligned}
$

The sum of three terms $=a+a r+a r^2+a=\left(1+r+r^2\right)$
$
\begin{aligned}
& =11(1+4+16) \\
& =11 \times 21=231
\end{aligned}
$

Hence, the required answer is 231.

Example 4: If $\mathrm{a}_1(>0), \mathrm{a}_2, \mathrm{a}_3, \mathrm{a}_4, \mathrm{a}_5$ are in a G.P., $\mathrm{a}_2+\mathrm{a}_4=2 \mathrm{a}_3+1$ and $3 \mathrm{a}_2+\mathrm{a}_3=2 \mathrm{a}_{4,}$, then $\mathrm{a}_2+\mathrm{a}_4+2 \mathrm{a}_5$ is equal to [JEE MAINS 2022]
Solution:
$
\begin{aligned}
& \mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_5 \longrightarrow \mathrm{G} \cdot \mathrm{p} \\
& \mathrm{a}_2+\mathrm{a}_4=2 \mathrm{a}_3+1 ; \quad 3 \mathrm{a}_2+\mathrm{a}_3=2 \mathrm{a}_4 \\
& \mathrm{ar}+\mathrm{ar}^3=2 \mathrm{a}^2+1 ; 3 \mathrm{ar}+\mathrm{ar}^2=2 \mathrm{ar}^3 \\
& 2 \mathrm{r}^2-\mathrm{r}-3=0 \\
& \mathrm{r}=-1 \cdot \frac{3}{2}
\end{aligned}
$

Now
$
\begin{aligned}
& \frac{3 \mathrm{a}}{2}+\frac{27 \mathrm{a}}{8}=\frac{3 \mathrm{a}}{2}+1 \\
& \therefore \quad \mathrm{n}=\frac{3}{2} \\
& \mathrm{a}=\frac{8}{3} \\
& \mathrm{a}_2+\mathrm{a}_4+2 \mathrm{a}_5=\mathrm{ar}+\mathrm{ax}^3+2 \mathrm{ax}^4 \\
& =40
\end{aligned}
$

Hence, the required answer is 40.

Example 5: Let $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3, \cdots$ be an increasing geometric progression of positive real numbers. If If $\mathrm{A}_1 \mathrm{~A}_3 \mathrm{~A}_5 \mathrm{~A}_7=\frac{1}{1296}$ and $\mathrm{A}_2+\mathrm{A}_4=\frac{7}{36}$, then the value of $\mathrm{A}_6+\mathrm{A}_8+\mathrm{A}_{10}$ is equal to [JEE MAINS 2022]

Solution:
$
\begin{aligned}
& \text { Let } \mathrm{A}_1=\mathrm{a} \& \text { common ratio=r. } \\
& \mathrm{a} \cdot \mathrm{ar}^2 \cdot \mathrm{ar}^4 \cdot \mathrm{ar}^6=\frac{1}{1296} \Rightarrow \mathrm{a}^4 \mathrm{r}^{12}=\frac{1}{1296} \Rightarrow \mathrm{a}^2 \mathrm{r}^6=\frac{1}{36} \Rightarrow \mathrm{ar}^2=\frac{1}{6} \\
& \Rightarrow \mathrm{A}_4=\frac{1}{6}, \quad \mathrm{~A}_2+\mathrm{A}_4=\frac{7}{36} \\
& \Rightarrow \mathrm{A}_2=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}=\mathrm{ar} \Rightarrow \mathrm{r}^2=6 \Rightarrow \mathrm{r}=\sqrt{6}, \mathrm{a}=\frac{1}{36 \sqrt{6}} \\
& A_6+A_8+A_{10}=\operatorname{ar}^5+\operatorname{ar}^7+\operatorname{ar}^9=\frac{1}{36 \sqrt{6}}\left(6^2 \sqrt{6}+6^3 \sqrt{6}+6^4 \sqrt{6}\right) \\
& =1+6+6^2=43 \\
&
\end{aligned}
$

Hence, the required answer is 43.