Geometric Progression

In Mathematics, Geometric Progression (G. P) is a sequence of non-zero numbers. Each of the succeeding terms is generated by multiplying the previous term by a constant factor. This constant factor is called the common ratio. Thus, in a G.P the ratio of successive terms remains constant and can be determined by dividing any term by the term that comes immediately before it.

Thus, the sequence a, $\text{ ar },a{r}^2,a{r}^3,a{r}^4,...$ is a GP, with a as the first term and r as the common ratio.

Example:- 3, 6, 12, 24, 48, …

        a = 3, r = 2

1.0Geometric Progression Definition and General Form

A Geometric Progression (GP) is a sequence of numbers where every term after the first is derived by multiplying the preceding term by a constant factor known as the common ratio. The general form of a GP can be represented as $\mathrm{a},{\mathrm{ar}}^2,{\mathrm{ar}}^2,{\mathrm{ar}}^3,\dots ,{\mathrm{ar}}^{\mathrm{n}-1}$, where a as the first term, r as the common ratio, and n as the term number.

Example:- 1, 2, 4, 8, 16, …

    a = 1, r = 2

2.0Types of Geometric Progression

Geometric Progression (G.P) can be categorized into two types based on the nature and properties of the sequence. 

1. Finite Geometric Progression
2. Infinite Geometric Progression

Finite Geometric Progression

A finite geometric progression is a sequence that consists of a limited number of terms. The sequence has a clearly defined first term, common ratio, and a last term. This type is often encountered in problems where the total number of terms is known or can be easily determined. 

Example: 2, 6, 18, 54. In this sequence, the first term (a) is 2, the common ratio (r) is 3, and there are 4 terms.

Infinite Geometric Progression

An infinite geometric progression is a sequence that continues indefinitely without terminating. This type of GP is particularly interesting in cases where the common ratio is a fraction (i.e., |r| < 1), leading to a convergent series.

Example: $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$ In this sequence, the first term (a) is $\frac{1}{2}$, and the common ratio (r) is $\frac{1}{2}.$

3.0Geometric Progression Formulas

1.  N^th Term of a Geometric Progression

$n^{th}term;{\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}$

2. Sum of the first n terms of a Geometric Progression

Sum of the first n terms; $S_n=\frac{a\left(r^n-1\right)}{r-1},\text{ if }r\ne 1$

3. Sum of Infinite Terms of a Geometric Progression

Sum of infinite G.P., $S_{\infty }=\frac{a}{1-r};0<|r|<1$

4.0Properties or Geometric Progression

• If each term of a G.P. is multiplied or divided by a non-zero quantity, the resulting sequence remains a G.P.
• Three consecutive terms of a GP :  a / r, a, a r;  
• Four consecutive terms of a GP : $a/r^3,a/r,ar,a{r}^3$ & so on. 
• If a, b, c are in G.P. then b^2=a c. $\Rightarrow T_k\cdot T_{n-k+1}=constant=a\cdot l$
• In a G.P., the product of any 2 terms equidistant from the beginning and the end is constant and equal to the product of the first and last terms.
• If every term of a G.P. be raised to the same power, then the resulting sequence is also a G.P.
• In a G.P., $T_r^2=T_{r-k}\cdot T_{r+k},k<r,r\ne 1$
• If the terms of a given G.P. are selected at regular intervals, the resulting sequence will also form a G.P.
• If $a_1,a_2,a_3\dots a_n$ is a G.P. of positive terms, then $\log a_1,\log a_2,\dots ..\log a_n$ is an A.P. and vice-versa.
• If $a_1,a_2,a_3\dots .$ and $b_1,b_2,b_3\dots ..$ are two G.P.'s then $a_1{b}_1{a}_2{b}_2,a_3{b}_3\dots .$ & $\frac{a_1}{b_1},\frac{a_2}{b_2},\frac{a_3}{b_3}\dots \dots \dots$ is also in G.P.

5.0Solved Examples of Geometric Progression

Example 1: If a, b, c, d and p are distinct real numbers such that

$\left(a^2+b^2+c^2\right)p^2-2p(ab+bc+cd)+\left(b^2+c^2+d^2\right)\le 0$ then a, b, c, d are in 

(A) A.P. (B) G.P. (C) H.P. (D) None of these

Ans. (B)

Solution:

Here, the given condition

$\left(a^2+b^2+c^2\right)p^2-2p(ab+bc+cd)+b^2+c^2+d^2\le 0$

$\Rightarrow (ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2\le 0$

⇒ square cannot be negative

$\therefore ap-b=0,bp-c=0,cp-d=0\Rightarrow p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\Rightarrow a,b,c,d$ are in G.P.

Example 2: A three-digit number has digits that form a geometric progression (G.P.). The sum of the first and third digits exceeds twice the middle digit by 1, and the sum of the first and second digits is two-thirds of the sum of the second and third digits. Find the number.

Solution:

Let the three digits be $a,ar\text{ and }a{r}^2$ then number is

$100a+10ar+a{r}^2$ ...(i)

Given, $a+a{r}^2=2ar+1$

$\Rightarrow a\left(r^2-2r+1\right)=1$

$\Rightarrow a(r-1{)}^2=1$ ...(ii)

Also given $a+ar=\frac{2}{3}\left(ar+a{r}^2\right)$

$\Rightarrow 3+3r=2r+2r^2\Rightarrow 2r^2-r-3=0\Rightarrow (r+1)(2r-3)$=0

$\therefore r=-1,3/2$

for r=-1, $a=\frac{1}{(r-1{)}^2}=\frac{1}{4}\notin I\therefore r\ne -1$

for $\begin{array}{rl}& r=3/2,\end{array}a=\frac{1}{{\left(\frac{3}{2}-1\right)}^2}=4$     {from (ii)}

From (i), the number is $100(4)+10(4)\left(\frac{3}{2}\right)+4{\left(\frac{3}{2}\right)}^2$

= 400 + 60 + 9 

= 469.

Example 3: If positive real numbers a, b, c are in G.P., then the equations a $x^2+2bx+c=0$ and $d{x}^2+2ex+f=0$ have a common root if $\frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in -

(A) A.P. (B) G.P. (C) H.P. (D) None of these

Ans. (A)

Solution:

a, b, c are in G.P $\Rightarrow b^2=ac$

Now the equation $a{x}^2+2bx+c=0$ can be rewritten as a $x^2+2\sqrt{ac}x+c=0$

$\Rightarrow (\sqrt{a}x+\sqrt{c}{)}^2=0\Rightarrow x=-\sqrt{\frac{c}{a}},-\sqrt{\frac{c}{a}}$

If the two given equations have a common root, then this root must be -$\sqrt{\frac{c}{a}}$ .

Thus $d\frac{c}{a}-2e\sqrt{\frac{c}{a}}+f=0\Rightarrow \frac{d}{a}+\frac{f}{c}=\frac{2e}{c}\sqrt{\frac{c}{a}}=\frac{2e}{\sqrt{ac}}=\frac{2e}{b}\Rightarrow \frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in A.P.

Example 4: Find the value of $0.32\overline{58}$

Solution:

Let R=0.32

$\overline{58}\Rightarrow R=0.32585858\dots$ . ...(i)

Here the number of figures which are not recurring is 2 and the number of figures which are recurring is also 2.

then 100 R = 32.585858...... ...(ii)

and 10000 R = 3258.5858..... ...(iii)

Subtracting (ii) from (iii) , we get  

9900 R=3226 $\Rightarrow R=\frac{1613}{4950}$

Alter Method: R = .32 + .0058 + .000058 + .00000058 +...........

$=.32+\frac{58}{10^4}\left(1+\frac{1}{10^2}+\frac{1}{10^4}+\dots \dots ..\infty \right)=.32+\frac{58}{10^4}\left(\frac{1}{1-\frac{1}{100}}\right)$

$=\frac{32}{100}+\frac{58}{9900}=\frac{3168+58}{9900}=\frac{3226}{9900}=\frac{1613}{4950}$

6.0Practice Problems Based of Geometric Progression

1. If the third term of G.P. is 4, then find the product of the first five terms.

(A) $4^4$

(B) $4^5$

(C) $2^7$

(D) $2^9$

Ans: B

2. Find the sum of the series $\frac{3}{4},1\frac{1}{2},3,\dots$ to 8 terms.

(A) 191

(B) $191\frac{1}{2}$

(C) $191\frac{1}{4}$

(D) None of these

Ans: C

3. Find a three-digit number whose consecutive digits form a G.P. If we subtract 792 from this number, the resulting number has the same digits in reverse order. Now, if we increase the second digit of the required number by 2, then the resulting digits will form an A.P.

(A) 931 (B) 842 (C) 421 (D) None of these

Ans: A

4. The sum of the first 6 terms of a G.P. is 9 times the sum of the first 3 terms; find the common ratio.

(A) 1 (B) 2 (C) 3 (D) 4

Ans: B

5. The sum of an infinite number of terms of a G.P. is 4, and the sum of their cubes is 192; find the series. 

Ans: $6,-3,1\frac{1}{2},\dots \text{ or }12,-24,48,\dots$

7.0Sample Questions on Geometric Progression

1. What is Geometric Progression (GP)?

Ans: A Geometric Progression (GP) is a sequence of numbers where every term after the first is derived by multiplying the preceding term by a constant factor known as the common ratio. The general form of a GP can be represented as $\mathrm{a},\mathrm{ar},{\mathrm{ar}}^2,{\mathrm{ar}}^3,\dots ,{\mathrm{ar}}^{\mathrm{n}-1}$, where a as the first term, r as the common ratio, and n as the term number.

2. How is the nth term of a GP calculated?

Ans: The nth term (T_n) of a GP can be calculated using the formula:

${\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}$, where a as the first term, r as the common ratio, and n is the term number.

3. What is the common ratio in a GP?

Ans: The common ratio (r) in a GP is the constant factor by which each term is multiplied to obtain the next term. It can be determined by dividing any term by its preceding term:

$r=\frac{T_{n+1}}{T_n}$

4. What is the sum of an infinite GP?

Ans: If the common ratio 0 < |r| < 1, the sum (S) of an infinite GP is: $S_{\infty }=\frac{a}{1-r}$

5. How do you determine if a sequence is a GP?

Ans: A sequence is a GP if the ratio of successive terms is constant. Mathematically, if $\frac{T_{n+1}}{T_n}=\frac{T_{n+2}}{T_{n+1}}$ for all n, then the sequence is a GP.