`star` Geometric Progression (G . P.)

`star` General term of a G .P.

`star` Sum to n terms of a G .P.

`star` Relationship Between A.M. and G.M.

`star` General term of a G .P.

`star` Sum to n terms of a G .P.

`star` Relationship Between A.M. and G.M.

`\color{fuchsia} \(★ mathbf\ul"Geometric Progression (G . P.)")`

A sequence `color(blue)(a_1, a_2, a_3, …, a_n, … )` is called geometric progression,

if each term is non-zero and `color(red)(a_(k+1)/a_k = r)` (constant ratio ), for `color(red)(k ≥ 1.)`

`\color{red} ✍️` By letting `color(blue)(a_1 = a)`, we obtain a geometric progression, `color(red)(a, ar, ar^2, ar^3,…...,)`

where `a` is called the `color(blue)"first term"` and `r` is called the `color(blue)"common ratio"` of the G.P.

We shall use the following notations with these formulae:

`color(red)(a) = ` the first term, ` \ \ color(red)(r) = `the common ratio, ` \ \ color(red)(l) =` the last term,

`color(red)(n) =` the numbers of terms,

`color(red)(S_n) =` the sum of n terms.

`color(red)("Let us consider the following sequences:")`

`(i) color(blue)(2,4,8,16,.............,) (ii) color(blue)(1/9 , -1/ 27, 1/81 , -1/243 ..............) (iii) color(blue)(.01,.0001,.000001,..................)`

In `(i)` we have `a_1=2 , a_2/a_1=2, a_3/a_2=2, a_4/a_3=2` and so on. In `(ii)`, we observe, `a_1=1/9 , a_2/a_1=-1/3, a_3/a_2=-1/3 , a_4/a_3=-1/3` and so on.

Similarly, state how do the terms in `(iii)` progress

In `(i),` this constant ratio is `2;` in `(ii)` it is `-1/3` and in `(iii)` the constant ratio is `0.01.`

Common ratio in geometric progression `(i), (ii)` and `(iii)` above are `2, -1/3` and `0.01,` respectively.

Such sequences are called geometric sequence or geometric progression abbreviated as `G.P`

A sequence `color(blue)(a_1, a_2, a_3, …, a_n, … )` is called geometric progression,

if each term is non-zero and `color(red)(a_(k+1)/a_k = r)` (constant ratio ), for `color(red)(k ≥ 1.)`

`\color{red} ✍️` By letting `color(blue)(a_1 = a)`, we obtain a geometric progression, `color(red)(a, ar, ar^2, ar^3,…...,)`

where `a` is called the `color(blue)"first term"` and `r` is called the `color(blue)"common ratio"` of the G.P.

We shall use the following notations with these formulae:

`color(red)(a) = ` the first term, ` \ \ color(red)(r) = `the common ratio, ` \ \ color(red)(l) =` the last term,

`color(red)(n) =` the numbers of terms,

`color(red)(S_n) =` the sum of n terms.

`color(red)("Let us consider the following sequences:")`

`(i) color(blue)(2,4,8,16,.............,) (ii) color(blue)(1/9 , -1/ 27, 1/81 , -1/243 ..............) (iii) color(blue)(.01,.0001,.000001,..................)`

In `(i)` we have `a_1=2 , a_2/a_1=2, a_3/a_2=2, a_4/a_3=2` and so on. In `(ii)`, we observe, `a_1=1/9 , a_2/a_1=-1/3, a_3/a_2=-1/3 , a_4/a_3=-1/3` and so on.

Similarly, state how do the terms in `(iii)` progress

In `(i),` this constant ratio is `2;` in `(ii)` it is `-1/3` and in `(iii)` the constant ratio is `0.01.`

Common ratio in geometric progression `(i), (ii)` and `(iii)` above are `2, -1/3` and `0.01,` respectively.

Such sequences are called geometric sequence or geometric progression abbreviated as `G.P`

Let us consider a `G.P. `with first non-zero term `‘a’` and common ratio `‘r’.`

The second term is obtained by multiplying `a` by `r,` thus `color(blue)(a_2 = ar).`

Similarly, third term is obtained by multiplying `a_2` by `r.` Thus, `color(blue)(a_3 = a_2r = ar^2),` and so on.

We write below these and few more terms.

1st term `color(blue)(= a_1 = a = ar^(1–1)),`

2nd term `color(blue)(= a_2 = ar = ar^(2–1)),`

3rd term `color(blue)(= a_3 = ar^2 = ar^(3–1))`,

4th term `color(blue)(= a_4 = ar^3 = ar^(4–1)),`

5th term `color(blue)(= a_5 = ar^4 = ar^(5–1))`,

Do you see a pattern? What will be `16^(th)` term?

`color(blue)(a_16 = ar^(16–1) = ar^15)`

Therefore, the pattern suggests that the `n^(th)` term of a G.P. is given by `color(red)(a _n =ar^( n−1).)`

Thus, a, G.P. can be written as `color(blue)(a, ar, ar^2, ar^3, … ar^(n – 1)) \ \ \ \ \; color(red)(a, ar, ar^2,...,ar^(n – 1)....... ;)` according as G.P. is finite or infinite, respectively.

The series `color(blue)(a + ar + ar^2 + ... + ar^(n–1))` or `color(red)(a + ar + ar^2 + ... + ar^(n–1) +....... )` are called finite or infinite geometric series, respectively.

The second term is obtained by multiplying `a` by `r,` thus `color(blue)(a_2 = ar).`

Similarly, third term is obtained by multiplying `a_2` by `r.` Thus, `color(blue)(a_3 = a_2r = ar^2),` and so on.

We write below these and few more terms.

1st term `color(blue)(= a_1 = a = ar^(1–1)),`

2nd term `color(blue)(= a_2 = ar = ar^(2–1)),`

3rd term `color(blue)(= a_3 = ar^2 = ar^(3–1))`,

4th term `color(blue)(= a_4 = ar^3 = ar^(4–1)),`

5th term `color(blue)(= a_5 = ar^4 = ar^(5–1))`,

Do you see a pattern? What will be `16^(th)` term?

`color(blue)(a_16 = ar^(16–1) = ar^15)`

Therefore, the pattern suggests that the `n^(th)` term of a G.P. is given by `color(red)(a _n =ar^( n−1).)`

Thus, a, G.P. can be written as `color(blue)(a, ar, ar^2, ar^3, … ar^(n – 1)) \ \ \ \ \; color(red)(a, ar, ar^2,...,ar^(n – 1)....... ;)` according as G.P. is finite or infinite, respectively.

The series `color(blue)(a + ar + ar^2 + ... + ar^(n–1))` or `color(red)(a + ar + ar^2 + ... + ar^(n–1) +....... )` are called finite or infinite geometric series, respectively.

Let the first term of a G.P. be `a` and the common ratio be `r.`

Let us denote by `S_n` the sum to first `n` terms of `G.P.` Then

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(S_n = a + ar + ar^2 +...+ ar^(n–1)) .................... (1)`

`color(red)(Case 1)` If `color(blue)(r = 1),` we have `color(blue)(S_n = a + a + a + ... + a)` (n terms) `color(red)(= na)`

`color(red)(Case 2)` If `color(blue)(r ≠ 1),` multiplying (1) by `r,` we have

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(rS_n = ar + ar^2 + ar^3 + ... + ar^n) ........................ (2)`

Subtracting (2) from (1), we get `(1 – r) S_n = a – ar^n = a(1 – r^n)`

`color(red)(S_n= (a(1 – r^n))/(1-r))` or `color(blue)(S_n= (a(r^n-1))/(r-1))`

Let us denote by `S_n` the sum to first `n` terms of `G.P.` Then

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(S_n = a + ar + ar^2 +...+ ar^(n–1)) .................... (1)`

`color(red)(Case 1)` If `color(blue)(r = 1),` we have `color(blue)(S_n = a + a + a + ... + a)` (n terms) `color(red)(= na)`

`color(red)(Case 2)` If `color(blue)(r ≠ 1),` multiplying (1) by `r,` we have

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(rS_n = ar + ar^2 + ar^3 + ... + ar^n) ........................ (2)`

Subtracting (2) from (1), we get `(1 – r) S_n = a – ar^n = a(1 – r^n)`

`color(red)(S_n= (a(1 – r^n))/(1-r))` or `color(blue)(S_n= (a(r^n-1))/(r-1))`

Q 3059812714

Find the 10th and nth terms of the G.P. `5, 25,125,… .`

Here `a = 5` and `r = 5. `Thus, `a_(10) = 5(5)^(10–1) = 5(5)^9 = 5^(10)`

and `a_n = ar^(n–1) = 5(5)^(n–1) = 5^n`

Q 3029012811

Which term of the G.P., 2,8,32, ... up to n terms is 131072?

Let `131072` be the nth term of the given G.P. Here `a = 2` and `r = 4.`

Therefore `131072 = a_n = 2(4)^(n – 1)` or `65536 = 4^(n – 1)`

This gives `48 = 4^(n – 1).`

So that `n – 1 = 8, i.e., n = 9.` Hence, 131072 is the 9th term of the G.P.

Q 3069012815

In a G.P., the 3rd term is 24 and the 6th term is 192.Find the 10th term.

Here,`a_3 = ar^2 = 24` ..............(1)

and `a_6 = ar^5 = 192` .................(2)

Dividing (2) by (1), we get r = 2. Substituting `r = 2` in (1), we get `a = 6`

Hence `a_(10) = 6 (2)^9 = 3072`

Q 3069112915

Find the sum of first n terms and the sum of first 5 terms of the geometric series `1+2/3+4/9 +.........`

Here `a = 1` and `r = 2/3` Therefore `S_n = (a (1-r^n))/(1-r) = ([1-2/3])/(1-2/3) = 3 [1-(2/3)^n]`

In particular `S_5 = 3 [ 1-(2/3)^5] = 3 xx (211)/(243) = (211)/(81)`

Q 3019112919

How many terms of the G.P. `3 , 3/2 , 3/4 , .............` are needed to give the sum `(3069)/(512)`

Let n be the number of terms needed. Given that `a = 3, r =1/2` and `S_n = (3069)/(512)`

Since `S_n = (a (1-r^n))/(1-r)`

Therefore `(3069)/(512) = ( 3 ( 1-1/(2^n)))/(1-1/2) = 6 (1-1/(2^n))`

`(3069)/(3072) = (1-1/(2^n))`

or `1/(2^n) = 1-3069/3072 = 3/3072 = 1/1024`

or `2^n = 1024 = 2^n = 2^(10)`

which gives n = 10.

Q 3029223111

The sum of first three terms of a G.P. is `13/12` and their product is `– 1.` Find the common ratio and the terms.

Let `a/r , a , ar ` be the first three terms of the G.P. Then

`a/r +ar +a = 13/12` ..............(1)

and `(a/r) (a) (ar) = 1` .................(2)

From (2), we get `a_3 = – 1,` i.e., `a = – 1` (considering only real roots)

Substituting `a = –1` in (1), we have

` -1/r -1-r = 13/12` or `12r^2+25r+12 = 0`

This is a quadratic in r, solving, we get `r= -3/4 ` or `-4/3`

Thus, the three terms of G.P. are `4/3 , -1 , 3/4` for `r = (-3)/4` and `r = (-3)/4` and `3/4 , -1 , 4/3` for `r = (-4)/3`

Q 3069223115

Find the sum of the sequence 7, 77, 777, 7777, ... to n terms.

This is not a G.P., however, we can relate it to a G.P. by writing the terms as

`S_n = 7 + 77 + 777 + 7777 + ... ` to n terms

` = 7/9 [ 9+99+999+9999+........+ to n term ]`

` = 7/9 [ (10-1) + (10^2-1) +(10^3-1) +(10^4-1) +......+ n terms]`

` = 7/9 [ (10+10^2+10^3 + ............. n terms ) - (1+1+1+......n terms) ]`

` = 7/9 [ (10(10^n -1))/(10-1) -n] =7/9 [ (10(10^n-1))/9-n]`

Q 3079323216

A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own.

Here `a = 2, r = 2` and `n = 10`

Using the sum formula `S_n = ( a (r^n -1))/(r-1)`

we have `S_(10) = 2(2^(10)-1) = 2046`

Hence, the number of ancestors preceding the person is 2046.

`\color{fuchsia} \(★ mathbf\ul"Geometric Mean (G .M.) ")`

The geometric mean of two positive numbers `a` and `b` is the number `color(red)(sqrt(ab )).`

Therefore, the geometric mean of `2` and `8` is `4.`

We observe that the three numbers `2,4,8` are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers.

Given any two positive numbers a and b, we can insert as many numbers as we like between them to make the resulting sequence in a G.P.

Let `color(blue)(G_1, G_2,…, G_n)` be `n` numbers between positive numbers `a` and `b` such that `color(blue)(a,G_1,G_2,G_3,…,G_n,b)` is a G.P.

Thus, `b` being the `(n + 2)th` term,we have `color(blue)(b = ar^(n + 1))` , or `color(blue)(r=(b/a)^(1/(n+1)))`

Hence `color(blue)(G_1= ar=a(b/a)^(1/(n+1)))`

`color(blue)(G_2= ar^2=a(b/a)^(2/(n+1)))`

`color(blue)(G_3= ar^3=a(b/a)^(3/(n+1)))`

`color(blue)(G_n=ar^n=a(b/a)^(n/(n+1)))`

The geometric mean of two positive numbers `a` and `b` is the number `color(red)(sqrt(ab )).`

Therefore, the geometric mean of `2` and `8` is `4.`

We observe that the three numbers `2,4,8` are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers.

Given any two positive numbers a and b, we can insert as many numbers as we like between them to make the resulting sequence in a G.P.

Let `color(blue)(G_1, G_2,…, G_n)` be `n` numbers between positive numbers `a` and `b` such that `color(blue)(a,G_1,G_2,G_3,…,G_n,b)` is a G.P.

Thus, `b` being the `(n + 2)th` term,we have `color(blue)(b = ar^(n + 1))` , or `color(blue)(r=(b/a)^(1/(n+1)))`

Hence `color(blue)(G_1= ar=a(b/a)^(1/(n+1)))`

`color(blue)(G_2= ar^2=a(b/a)^(2/(n+1)))`

`color(blue)(G_3= ar^3=a(b/a)^(3/(n+1)))`

`color(blue)(G_n=ar^n=a(b/a)^(n/(n+1)))`

Q 3079423316

Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.

Let `G_1, G_2,G_3` be three numbers between 1 and 256 such that `1, G_1,G_2,G_3 ,256` is a G.P.

Therefore `256 = r^4` giving `r = ± 4` (Taking real roots only)

For `r = 4,` we have `G_1 = ar = 4, G_2 = ar^2 = 16, G_3 = ar^3 = 64`

Similarly, for `r = – 4,` numbers are `– 4,16 `and `– 64.`

Hence, we can insert, `4, 16, 64` or `– 4, 16, –64,` between 1 and 256 so that the resulting

sequences are in G.P.

Let `A` and `G` be A.M. and G.M. of two given positive real numbers `a` and `b,` respectively.

Then `color(blue)(A= (a+b)/2)` and `color(blue)(G=sqrt(ab))`

Thus, we have `color(blue)(A – G)=(a+b)/2-sqrt(ab)= (a+b-2sqrt(ab))/2`

` \ \ \ \ \ \ \ \ \ \ \ color(blue)(A – G)=(sqrt(a)-sqrt(b))^2/2 >= 0`

we obtain the relationship `color(red)(A ≥ G.)`

Then `color(blue)(A= (a+b)/2)` and `color(blue)(G=sqrt(ab))`

Thus, we have `color(blue)(A – G)=(a+b)/2-sqrt(ab)= (a+b-2sqrt(ab))/2`

` \ \ \ \ \ \ \ \ \ \ \ color(blue)(A – G)=(sqrt(a)-sqrt(b))^2/2 >= 0`

we obtain the relationship `color(red)(A ≥ G.)`

Q 3019523419

If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers.

Given that `A. M. = (a+b)/2 = 10` ................(1)

and `G. M. = sqrt(ab) = 8` ........(2)

From (1) and (2), we get

`a + b = 20` ... (3)

`ab = 64` ... (4)

Putting the value of a and b from (3), (4) in the identity `(a – b)^2 = (a + b)^2 – 4ab,` we get

`(a-b)^2 = 400-256 = 144`

`a-b = pm12 ` .........(5)

Solving (3) and (5), we obtain

`a = 4, b = 16 or a = 16, b = 4`

Thus, the numbers a and b are 4, 16 or 16, 4 respectively.