Progressions-A.P. and G.P. (Part-4)

© ® Written by Sankar Ghosh

Hello friends!

I hope you are doing well.

Over the last three blog posts, we explored Arithmetic Progression (A.P.) in depth — from its definition and general term to the sum of n terms, properties, and even special series. Each article was filled with detailed explanations, examples, and exercises to make learning easier.

Now, it’s time to begin a new journey! 

In today’s post, we enter into the world of Geometric Progression (G.P.). We’ll understand its basics, explore its general term, middle term and sum to `n` terms and gradually move towards problem-solving.

As always, I look forward to your valuable comments and suggestions. Your feedback helps me improve and keeps me motivated to share more with you!


Introduction: Geometric Progression (G.P.)

In the last three blog post, we studied Arithmetic Progression (A.P.), where the difference between two consecutive terms is always the same. For example:

1, 3, 5, 7, … → here the common difference is 2.

But not all sequences follow this rule. Let’s look at another sequence:

1, 2, 4, 8, 16, …

Here, the difference between consecutive terms is not constant:

2 – 1 = 1, 4 – 2 = 2, 8 – 4 = 4, 16 – 8 = 8.

So, this is not an A.P. But notice something interesting:

  • Each term is obtained by multiplying the previous term by 2.
  • The ratio of any term to its previous term is always 2.

Such type of sequence is called a Geometric Progression (G.P.).

Definition of G.P.:

A sequence is called a Geometric Progression if the ratio of any term to its preceding term is always a constant number (Except 0 or 1).

This constant number is known as the common ratio (r).

What you will learn in this chapter:

By the end of this chapter, you will be able to:

  • Understand the definition and properties of G.P.
  • Find the general term and the nth term of a G.P.
  • Derive and calculate the sum of n terms of a G.P. in different cases.
  • Solve problems based on special types of G.P.
  • Apply the concepts to real examples and simple problem-solving.

In the upcoming sections, we’ll explore G.P. step by step, with examples and exercises, just like we did for A.P.

Like Arithmetic Progression, the numbers by which a Geometric Progression is formed, each are called terms of the G.P. and starting from the beginning the terms are denoted by first term`\left(t_1\right)` , second term`\left(t_2\right)`, third term`\left(t_3\right)`,...respectively. In this way the `n`th term`\left(t_n\right)` obtain thus is called general term and by putting `n=1,2,3,...`in this term we can determine the different terms of the G.P.

Let us consider a G.P. whose first term is `a` and common ratio is `r`. Then from the beginning if the terms are denoted by `t_1,t_2,t_3,...` then we can write `t_1=a,t_2=t_1\.r=a\.r,t_3=t_2\.r=ar\.r=ar^2,t_4=t_3\.r=ar^2\.r=ar^3` and in this way we can write `t_n=ar^{n-1}`. This is the general formula of `n`th term of G.P.

Remark: (i) Here, we see that `n`th term `t_n` is dependent only on `a` and `r`. Thus, if we know the values of `a` and `r` we can determine completely the Geometric Progression.

(ii) If `a,b,c` are in G.P., then `\frac{b}{a}=\frac{c}{b}\impliesb^2=ac`.

(iii) If three numbers are in G.P. we consider them as `\frac{a}{r},a,ar` and for four terms we take, `\frac{a}{r^3},\frac{a}{r},ar,ar^3` for the sake of convenience.


ILLUSTRATIVE EXAMPLES


Example 1: Find the G.P. whose first term is 2 and common difference is 3.

Solution: Given that the first term `a=2` and common ratio `r=3` `\thereforet_n=ar^{n-1}=2\.3^{n-1}`.

Now, putting `n=1,2,3,...` we get `t_1=2\.3^{1-1}=2,t_2=2\.3^{2-1}=6,t_3=2\.3^{3-1}=18,...`

Hence, the required G.P. is `2,6,18,....`

Example 2: Find the `20`th and `n`th term of the G.P., `\frac{5}{2},\frac{5}{4},\frac{5}{8},...`

Solution: Here, the first term `a=\frac{5}{2}` and the common ratio `r=\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2}`

`\therefore`        `t_20=ar^{20-1}=\frac{5}{2}\left(\frac{1}{2}\right)^19=\frac{5}{2^20}` and `t_n=\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}=\frac{5}{2^n}`.

Example 3: Find the 12th term of a G.P. whose `8`th term is `192` and the common ration is `2`

Solution: We are given that, `t_8` is `192` and common ratio is `2`.  Late the first term is `a` and common ratio is `r`

`\therefore`        `t_8=192` and `r=2\impliesar^7=192\impliesa2^7=192\impliesa=frac{192}{128}=\frac{3}{2}`.

`\therefore`        `t_12=ar^11=\frac{3}{2}2^11=3\.2^10=3072`.

Example 4: Which term of the G.P. `\sqrt3,3,3\sqrt3,...` is `729`? 

Solution: Given that first term `a=\sqrt3` and common ratio `r=\frac{3\sqrt3}{3}=\sqrt3`. 

Let the `n`th term of the G.P. be `729`    `\thereforear^{n-1}=\sqrt3\left(\sqrt3\right)^{n-1}=729\implies\left(\sqrt3\right)^n=729\implies3^\frac{n}{2}=3^6\impliesn=12`.

Hence, `12`th term of the G.P. is `729`

Example 5: If the second term and fifth term of a G.P. is `9` and `243` respectively then find the first term and fourth term of the progression.

Solution: Let the first term and the common ratio of the progression be `a` and `r` respectively.

`\therefore`        general term`\left(t_n\right)=a\.r^{n-1}`. 

Now, according to the question, `t_2=a\.r^{2-1}=ar=9` and `t_5=a\.r^{5-1}=ar^4=243`.

Now, `\frac{t_5}{t_2}=\frac{243}{9}\implies\frac{ar^4}{ar}=\frac{243}{9}` [since `rne0`] `\impliesr^3=27\impliesr=3`

`\therefore`        `ar=9\impliesa=3` `\therefore`    `t_n=3\.3^{n-1}=3^n`.

Hence,  `t_4=3^4=81`.   

Example 6: If `x,y,z` be the `p`th,`q`th and `r`th terms respectively both of an A.P. and of a G.P., then prove that, `x^{y-z}\.y^{z-x}\.z^{x-y}=1`  

Solution:  Let the first term and common difference of the A.P. be `a` and `d` respectively and the first of the G.P. be `m` and the common ratio be `n`

Now, according to the problem, we have for the A.P., `t_p=a+\left(p-1\right)d=x` ..........(1)

`t_q=a+\left(q-1\right)d=y`..............(2) and `t_r=a+\left(r-1\right)d=z`................(3) .

Now, (1)-(2) gives `{a+\left(p-1\right)d}-{a+\left(q-1\right)d}=x-y\implies\left(p-q\right)d=x-y`.........(4).

(2)-(3) & (3)-(1) gives `\left(q-r\right)d=y-z`........(5) and `\left(r-p\right)d=z-x`........(6) respectively.

For G.P. we have,    `t_p=mn^{p-1}=x,t_q=mn^{q-1}=y` and `t_r=mn^{r-1}=z`.

`\therefore`    `x^{y-z}\.y^{z-x}\.z^{x-y}=\left(mn^{p-1}\right)^{\left(q-r\right)d}\.\left(mn^{q-1}\right)^{\left(r-p\right)d}\.\left(mn^{r-1}\right)^{\left(p-q\right)d}`

`\implies` `x^{y-z}\.y^{z-x}\.z^{x-y}`

`=m^{\left(q-r\right)d+\left(r-p\right)d+\left(p-q\right)d}\.n^{\left(p-1\right)\left(q-r\right)d+\left(q-1\right)\left(r-p\right)d+\left(r-1\right)\left(p-q\right)d}`

`=m^0\.n^0=1\times1=1`.

Example 7: The product of three consecutive terms of a G.P. is 27 and their sum is 13. Find those three numbers. 

Solution: Let the three numbers are `\frac{a}{r},a,ar` 

`\therefore`        `\frac{a}{r}\.a\.ar=27\impliesa^3=27\impliesa=3`.    Also we have,    `\frac{a}{r}+a+ar=13\impliesa\left(\frac{1}{r}+1+r\right)=13\implies\frac{3\left(1+r+r^2\right)}{r}=13`

`\implies3r^2-10r+3=0\implies\left(3r-1\right)\left(r-3\right)=0\impliesr=\frac{1}{3},3`.

Hence, the required three numbers are: `9,3,1` (When `a=3,r=\frac{1}{3}`) or `1,3,9` (When `a=3,r=3`).

Example 8: Sum of the four numbers in G.P. is 180; sum of their two extreme numbers is 108; find the numbers.

Solution: Let the four numbers are `\frac{a}{r^3},\frac{a}{r},ar,ar^3` 

Now, by the problem we have,

`\frac{a}{r^3}+\frac{a}{r}+ar+ar^3=180`.........(1) and `\frac{a}{r^3}+ar^3=108`.........(2)

Now, from (1) and (2) we get, `\frac{a}{r}+ar=72`.........(3). 

Now, dividing (2) by (3) we get,`\frac{r^3+\frac{1}{r^3}}{r+\frac{1}{r}}=\frac{108}{72}=\frac{3}{2}`

`\implies`    `\frac{\left(r+\frac{1}{r}\right)\left(r^2-1+\frac{1}{r^2}\right)}{\left(r+\frac{1}{r}\right)}=\frac{3}{2}\implies\frac{r^4-r^2+1}{r^2}=\frac{3}{2}`

`implies` `2r^4-5r^2+2=0\implies2r^4-4r^2-r^2+2=0\implies\left(2r^2-1\right)\left(r^2-2\right)=0`

`\impliesr^2=\frac{1}{2},2\impliesr=\frac{1}{\sqrt2},\sqrt2`.

Now, putting `r=\sqrt2` in equation (3), we get `a=24\sqrt2`. 

Hence the numbers are `12,24,48,96` (when `a=24\sqrt2,r=sqrt2`) 

Or                                      `96,48,34,12` (when `a=24\sqrt2,r=\frac{1}{\sqrt2}`).


 Exercise-2(A) 

(MCQ)

A. Chose thee correct Answer
  1. The third term of a G.P. is `4`; the product of them is:
    (a) `4^2`
    (b) `4^3`
    (c) `4^4`
    (d) `4^5`.
  2. How many terms are there in the series `4+8+16+..........+256`?
    (a) `6`
    (b) `7`
    (c) `8`
    (d) `9`.
  3. If `x,2y,3z` are in A.P., where the distinct numbers `x,y,z`are in G.P., then the common ratio of the G.P. is
    (a) `3`
    (b) `\frac{1}{3}`
    (c) `2`
    (d) `\frac{1}{2}`.
  4. The fifth term of the series `\frac{10}{9},\frac{1}{3}sqrt\frac{20}{3},\frac{2}{3},....` is
    (a) `\frac{1}{3}` 
    (b) `1`
    (c) `\frac{2}{5}`
    (d) `\sqrt\frac{2}{3}`.
  5. The first two terms of a G.P. add up to `12`. The sum of the third and the fourth term is `48`. If the terms of the G.P. are alternately positive and negative, then the first term is
    (a) `4`
    (b) `-4`
    (c) `12`
    (d) `-12`.
  6. If `x,2x+2,3x+3` are the first three terms of a G.P., then the fourth term is
    (a) `13.5`
    (b) `-13.5`
    (c) `-27`
    (d) `27`.
  7. If the `\left(m+n\right)`th term of a G.P. id `16` and `\left(m-n\right)`th term is `9`, then its `m`th term will be
    (a) `5`
    (b) '7`
    (c) `12`
    (d) `25`.
  8. If `x,y,z` are in geometric progression, then `\frac{1}{x^2-y^2}+\frac{1}{y^2}=`
    (a) `\frac{1}{y^2-x^2}`
    (b) `\frac{1}{y^2-z^2}`
    (c) `\frac{1}{z^2-y^2}`
    (d) `\frac{1}{z^2-x^2}`.
  9. If `a,b,c` are three unequal numbers such that `a,b,c` ar in A.P. and `a,b-a,c-a` ar in G.P., them `a:b:c` is
    (a) `1:2:3`
    (b)`1:3:5`
    (c) `1:4:7`
    (d) none of these.
  10. If `p`th, `q`th and `r`th terms of a G.P. are `x,y,z` then `x^{q-r}\.y^{r-p}\.z^{p-q}` equals
    (a) `0`
    (b) `1`
    (c) `xyz`
    (d) none of these. 

B. Answer the following questions

  1. Find the `10`th term and `p`th term of the progression `-6,2,-\frac{2}{3},....` 
    [Ans: `\frac{2}{3^8},\left(-1\right)^p\.\frac{2}{3^{p-2}}`]
  2. If the `4`th term and `9`th term of a geometric progression are respectively `\frac{1}{4}` and `\frac{1}{128}`. Determine the G.P.
    [Ans: `1,3,9,27,....`, `1,-3,9,-27,....`]
  3. If the fourth term of G.P. is nine times the second term and the seventh term is `729`, then determine the G.P.
    [Ans: `4,12,26,...`]
  4. If the `p`th term and `q`th term of a G.P. are `x` and `y` respectively, then show that its `\left(p+q\right)th` term is `\left(x^\frac{p}{p-q}\.y^\frac{q}{q-p}\right)`.
  5. Is `600` be a term of the progression `1,5,25,....` [Ans: No]
  6. For what value of `r` `2r,2r+1,2r+3` will be in G.P. [Ans: `r=\frac{1}{2}`]
  7. If the `6`th and `13`th term of a Geometric Progression are respectively `64` and `8192`, then find its `10`th term. [Ans: `1024`]
  8. If the `4`th, `10`th and `16`th terms of a G.P. are `x`, `y` and `z` respectively, prove that `x,y,z` are on G.P.
  9. The sum of three consecutive terms of an A.P. be 15. If `1,4,15` are added to those three terms then they form a G.P. Find the three numbers. [Ans: `2,5,8;26,5,-16`]
  10. Three rational numbers are in G.P. and their sum is `65` and the product of first and third numbers is 225. Find the common ratio of the G.P.  [Ans: `3` or `\frac{1}{3}`]
  11.  Sum of three numbers in G.P. is 21 and sum of their squares is 189. Find the numbers.[Ans: `3,6,12`]
  12.  If `x,y,z` are in G.P. and `p,q,r` in A.P., then show that `x^{q-r}\.y^{r-p}\.z^{p-q}=1`.
  13. If in a G.P. first term is `a`, `n`th term is `b`, and product of first `n` terms is `P`, then prove that, `P^2=\left(ab\right)^n`.  


    





 

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