11类RD Sharma解决方案–第20章几何级数-练习20.6

问题1：在27和1/81之间插入6个几何平均值。

解决方案：

Let the six geometric means be A₁, A₂, A₃, A₄, A₅, A₆.

Now, these 6 terms are to be added between 27 and 1/81.

So the G.P. becomes, 27, A₁, A₂, A₃, A₄, A₅, A₆,1/81 with first term(a) = 27, number of terms(n) = 8 and 8th term(a₈) = 1/81.

We know nth term of a G.P. is given by a_n = ar^n-1, where r is the common ratio.

=> a₈ = 1/81

=> 27(r)^8-1 = 1/81

=> r⁷ = 1/(81×27)

=> r⁷ = (1/3)⁷

=> r = 1/3

So, geometric means are:

A₁ = ar = 27(1/3) = 9

A₂ = ar² = 27(1/3)² = 3

A₃ = ar³ = 27(1/3)³ = 1

A₄ = ar⁴ = 27(1/3)⁴ = 1/3

A₅ = ar⁵ = 27(1/3)⁵ = 1/9

A₆ = ar⁶ = 27(1/3)⁶ = 1/27

Therefore, the 6 geometric means between 27 and 1/81 are 9, 3, 1, 1/3, 1/9, 1/27. 

问题2.在16和1/4之间插入5个几何平均值。

解决方案：

Let the five geometric means be A₁, A₂, A₃, A₄, A₅.

Now, these 5 terms are to be added between 16 and 1/4.

So the G.P. becomes, 16, A₁, A₂, A₃, A₄, A₅,1/4 with first term(a) = 16, number of terms(n) = 7 and 7th term(a₇) = 1/4.

We know nth term of a G.P. is given by a_n = ar^n-1, where r is the common ratio.

=> a₇ = 1/4

=> 16(r^7-1) = 1/4

=> r⁶ = 1/64

=> r⁶ = (1/2)⁶

=> r = 1/2

So, geometric means are:

A₁ = ar = 16(1/2) = 8

A₂ = ar² = 16(1/2)² = 4

A₃ = ar³ = 16(1/2)³ = 2

A₄ = ar⁴ = 16(1/2)⁴ = 1

A₅ = ar⁵ = 16(1/2)⁵ = 1/2

Therefore, the 5 geometric means between 16 and 1/4 are 8, 4, 2, 1, 1/2.

问题3.在32/9和81/2之间插入5个几何平均值。

解决方案：

Let the five geometric means be A₁, A₂, A₃, A₄, A₅.

Now, these 5 terms are to be added between 32/9 and 81/2.

So the G.P. becomes, 32/9, A₁, A₂, A₃, A₄, A₅, 81/2 with first term(a) = 32/9, number of terms(n) = 7 and 7th term(a₇) = 81/2.

We know nth term of a G.P. is given by a_n = ar^n-1, where r is the common ratio.

=> a₇ = 81/2

=> (32/9)(r^7-1) = 81/2

=> r⁶ = (81×9)/(2×32)

=> r⁶ = (3/2)⁶

=> r = 3/2

So, geometric means are:

A₁ = ar = (32/9)×(3/2) = 16/3

A₂ = ar² = (32/9)×(3/2)² = 8

A₃ = ar³ = (32/9)×(3/2)³ = 12

A₄ = ar⁴ = (32/9)×(3/2)⁴ = 18

A₅ = ar⁵ = (32/9)×(3/2)⁵ = 27

Therefore, the 5 geometric means between 32/9 and 81/2 are 16/3, 8, 12, 18, 27.

问题4.找到以下几对数字的几何均值：

(i)2和8

(ii)a ³ b和ab ³

(iii)–8和–2

解决方案：

We know geometric mean between two numbers, a and b is given by $\sqrt{ab}$.

(i) 2 and 8

Here, a = 2 and b = 8

So, G.M. = $\sqrt{2(8)}$

= $\sqrt{16}$

= 4

(ii) a³b and ab³

Here, a = a³b and b = ab³

So, G.M. = $\sqrt{a^3b×ab^3}$

= $\sqrt{a^4b^4}$

= a²b²

(iii) –8 and –2

Here, a = –8 and b = –2

G.M. = $\sqrt{(-8)×(-2)}$

= $\sqrt{16}$

= 4

问题5.如果a是GM的2和1/4，则找到a。

解决方案：

We know geometric mean between two numbers, a and b is given by $\sqrt{ab}$.

According to the question,

a = $\sqrt{2×(1/4)}$

= $\sqrt{1/2}$

= $\frac{1}{\sqrt{2}}$

Therefore, the value of a is $\frac{1}{\sqrt{2}}$.

问题6：找到两个数字，它们的AM为25，GM为20。

解决方案：

We know geometric mean between two numbers, a and b is given by $\sqrt{ab}$ and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

=> $\sqrt{ab}$ = 20 ……. (1)

And

=> (a+b)/2 = 25

=> a+b = 50

=> b = 50–a ……. (2)

From (1) and (2), we get,

=> $\sqrt{a(50-a)}$ = 20

Squaring both sides, we get,

=> a(50 –a) = 400

=> a² – 50a + 400 = 0

=> a²– 40a–10a+400 = 0

=> a(a– 40) – 10(a– 40) = 0

=> (a– 40) (a– 10) = 0

=> a = 40 or a = 10

Putting these in (2) we get,

When a = 40, then b = 10 and 

When a = 10, then b = 40.

Therefore, the numbers are 10 and 40.

问题7.在x中构造一个二次项，使得其根的AM为A，GM为G。

解决方案：

Suppose the roots of the quadratic equation are a and b.

We know geometric mean between two numbers, a and b is given by $\sqrt{ab}$ and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

A.M. of roots = (a+b)/2 = A

a + b = 2A ….. (1) 

And G.M. of roots = $\sqrt{ab}$ = G

ab = G²  … (2)

Now, we know a quadratic equation in x with roots a and b is given by,

x² – (a+b)x + (ab) = 0

From (1) and (2), we get,

x² – 2Ax + G² = 0 

Therefore, the required quadratic equation is x² – 2Ax + G² = 0.

问题8.两个数字的和是其几何平均值的6倍。证明数字在比例之内$(3+2\sqrt{2}):(3-2\sqrt{2})$

解决方案：

Let the two numbers be a and b. We know geometric mean between two numbers, a and b is given by $\sqrt[]{ab}$.

According to the question,

=> a+b = 6$\sqrt{(ab)}$

=> $\frac{a+b}{2\sqrt{(ab)}}$ = $\frac{3}{1}$

Applying Componendo and Dividendo on both sides, we get,

=> $\frac{a+b+2\sqrt{(ab)}}{a+b-2\sqrt{(ab)}}$ = $\frac{3+1}{3-1}$

=> $\left[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right]^2$ = $\frac{2}{1}$

=> $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$ = $\frac{\sqrt{2}}{1}$

By again applying Componendo and Dividendo on both sides, we get,

=> $\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}$ = $\frac{\sqrt{2}+1}{\sqrt{2}-1}$

=> $\frac{2\sqrt{a}}{2\sqrt{b}}$ = $\frac{\sqrt{2}+1}{\sqrt{2}-1}$

=> $\frac{\sqrt{a}}{\sqrt{b}}$ = $\frac{\sqrt{2}+1}{\sqrt{2}-1}$

Squaring both sides, we get

=> $\frac{a}{b}$ = $\left[\frac{\sqrt{2}+1}{\sqrt{2}-1}\right]^2$

=> $\frac{a}{b}$ = $\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$

=> $\frac{a}{b}$ = $\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$

Hence proved.

问题9.如果二次方程式的根的AM和GM分别为8和5，则求出二次方程式。

解决方案：

Suppose the roots of the quadratic equation are a and b.

We know geometric mean between two numbers, a and b is given by $\sqrt{ab}$ and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

A.M. of roots = (a+b)/2 = 8

a+b = 16 ….. (1)

And G.M. of roots = $\sqrt{ab}$ = 5

ab = 25  .… (2)

Now, let the quadratic equation with roots a and b is given by,

x² – (a+b)x + (ab) = 0

From (1) and (2), we get,

x² – 16x + 25 = 0

Therefore, the required quadratic equation is x² – 16x + 25 = 0.

问题10.两个正数a和b的AM和GM分别是10和8。查找数字。

解决方案：

We know geometric mean between two numbers, a and b is given by $\sqrt{ab}$ and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

=> $\sqrt{ab}$ = 8 ……. (1)

And,

=> (a+b)/2 = 10

=> a+b = 20

=> b = 20–a ……. (2)

From (1) and (2), we get,

=> $\sqrt{a(20-a)}$ = 8

Squaring both sides, we get,

=> a(20–a) = 64

=> a²–20a+64 = 0

=> a²–16a–4a+64 = 0

=> a(a–16) – 4(a–16) = 0

=> (a–4) (a–16) = 0

=> a = 4 or a = 16

Putting a = 4 in (2), we get b = 16. And,

Putting a = 16 in (2), we get b = 4.

Therefore, the numbers are 4 and 16.

问题11。证明两个量之间的n个几何平均值的乘积等于这两个量的几何平均值的n次幂。

解决方案：

Suppose we have a GP with first term a, common ratio r and number of terms n.

We have to add these n terms of GP between two quantities such that the GP remains maintained. So total number of terms become (n+2).

We know nth term of a G.P. is given by a_n = ar^n-1.

So, the last term of the GP ,i.e., (n+2)^th term will be, a_n+2 = ar^n+2-1 = arⁿ⁺¹  

We know geometric mean between two numbers, a and b is given by $\sqrt{ab}$.

The GM of a and arⁿ⁺¹ will be, G₁ = $\sqrt{a(ar^{n+1})}$ = $\left(a^2r^{n+1}\right)^{\frac{1}{2}}$  

Hence, L.H.S. = $G^n_1$ = $\left(a^2r^{n+1}\right)^{\frac{n}{2}}$

Now, R.H.S. = Product of n geometric means between these two quantities, G₂ = ar × ar² × . . . . × arⁿ

= $a^nr^{1+2+....+n}$

= $a^nr^{\frac{n(n+1)}{2}}$

= $\left[a^2r^{n+1}\right]^{\frac{n}{2}}$

= $G^n_1$

= L.H.S.

Hence, Proved.

问题12：如果两个正数a和b(a> b)的AM是其几何平均值的两倍。证明a：b = $(2+\sqrt{3}):(2-\sqrt{3})$

解决方案：

We know geometric mean between two numbers, a and b is given by $\sqrt{ab}$ and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

AM = 2(GM)

=> $\frac{a+b}{2}$ = $2\sqrt{ab}$

=> $\frac{a+b}{2\sqrt{(ab)}}$ = $\frac{2}{1}$

Applying Componendo and Dividendo on both sides, we get,

=> $\frac{a+b+2\sqrt{(ab)}}{a+b-2\sqrt{(ab)}}$ = $\frac{2+1}{2-1}$

=> $\left[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right]^2$ = $\frac{3}{1}$

=> $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$= $\frac{\sqrt{3}}{1}$

By again applying Componendo and Dividendo on both sides, we get,

=> $\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}$= $\frac{\sqrt{3}+1}{\sqrt{3}-1}$

=> $\frac{2\sqrt{a}}{2\sqrt{b}}$= $\frac{\sqrt{3}+1}{\sqrt{3}-1}$

=> $\frac{\sqrt{a}}{\sqrt{b}}$ = $\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Squaring both sides, we get

=> $\frac{a}{b}$ = $\left[\frac{\sqrt{3}+1}{\sqrt{3}-1}\right]^2$

=> $\frac{a}{b}$ = $\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}$

=> $\frac{a}{b}$ = $\frac{4+2\sqrt{3}}{4-2\sqrt{3}}$

=> $\frac{a}{b}$ = $\frac{2+\sqrt{3}}{2-\sqrt{3}}$

Hence, proved.

问题13.如果在任意两个正数之间插入一个AM，A和两个几何平均值G ₁和G _{2，则表明}$\frac{G^2_1}{G_2}+\frac{G^2_2}{G_1} = 2A$

解决方案：

Let the two positive numbers be a and b.

Now value of one AM between a and b, A = (a+b)/2.

So, 2A = a+b . . . . (1) 

If we add two geometric means between a and b, the GP becomes a,G₁,G₂,b.

Now, we know b = ar^4-1, where r is the common ratio.

=> r³ = $\frac{b}{a}$

=> r = $\left[\frac{b}{a}\right]^{\frac{1}{3}}$

So, G₁ = ar = $a\left[\frac{b}{a}\right]^{\frac{1}{3}}$ = $a^{\frac{2}{3}}b^{\frac{1}{3}}$

G₂ = ar² = $a\left[\frac{b}{a}\right]^{\frac{2}{3}}$ = $a^{\frac{1}{3}}b^{\frac{2}{3}}$

Now, L.H.S. = $\frac{G^2_1}{G_2}+\frac{G^2_2}{G_1}$

= $\frac{a^{\frac{4}{3}}b^{\frac{2}{3}}}{a^{\frac{1}{3}}b^{\frac{2}{3}}}+\frac{a^{\frac{2}{3}}b^{\frac{4}{3}}}{a^{\frac{2}{3}}b^{\frac{1}{3}}}$

= ab⁰ + a⁰b

= a+b

= 2A [From (1)]

= R.H.S.

Hence, proved.