11类RD Sharma解决方案–第20章几何级数-练习20.1 |套装1

问题1.证明接下来的每个进展都是GP。另外，在每种情况下找到共同的比例：

(i)4，-2，1，-1 / 2，……。

(ii)-2 / 3，-6，-54，…………。

(iii)a，3a ² /4，9a ³ /16，……。

(iv)1 / 2、1 / 3、2 / 9、4 / 27，……。

解决方案：

(i) Given a1 = 4, a2 = -2, a3 = 1

a2/a1 = -2/4 = -1/2

a3/a2 = -1/2

a4/a1 = (-1/2)/1 =-1/2

Now, a2/a1 = a3/a2 = a4/a3 = -1/2

Thus, a1, a2, a3, a4,… are in GP and common ratio(r) is -1/2.

(ii) Given a1 = -2/3, a2 = -6, a3 = -54

a2/a1 = -6/(-2/3) = 9

a3/a2 = -54/-6 = 9

Now, a2/a1 = a3/a2 = 9

Thus, a1, a2, a3,…… are in GP and common ratio(r) is 9.

(iii) Given a1= a, a2= 3a²/4, a3 = 9a³/16

a2/a1 = (3a²/4)/a = 3a/4

a3/a2 = (9a³/16)/(3a²/4) = 3a/4

 Now, a2/a1 = a3/a2 = 9

Thus, a1, a2, a3,…… are in GP and common ratio(r) is 3a/4.

(iv) Given a1=1/2, a2=1/3, a3=2/9

a2/a1 = (1/3)/(1/2) = 2/3

a3/a2 = (2/9)/(1/3) = 2/3

Now, a2/a1 = a3/a2 = 2/3

Thus, a1, a2, a3,…… are in GP and common ratio(r) is 2/3.

问题2。证明由an = 2 /(3 ⁿ )定义的序列是nGP。

解决方案：

Given an = 2/(3ⁿ)

We put n=1,2,3,….., n∈N 

The sequence is 2/3, 2/(3²), 2/(3³), ………..

a2/a1 = (2/(3²))/(2/3)=1/3

a3/a2 = (2/(3³))/(2/(3²)) = 1/3

Now, a2/a1 = a3/a2 = 1/3

Thus, a1,a2,a3,…… are in GP and common ratio(r) is 1/3.

Hence, the given sequence is in GP.

问题3.查找：

(i)GP 1,4,16,64，……的第9个任期。

(ii)GP的第十个任期-3 / 4、1 / 2，-1 / 3，-2 / 9，…………。

(iii)GP 0.3、0.06、0.012，…………..的第八项。

(iv)GP 1 /(a ³ x ³ )，ax，a5x ⁵ ，…..的第十二项。

(v)GP√3,1/√3,1/3√3，………。的第n个项。

解决方案：

(i) Given a2/a1 = 4/1 = 4

a3/a2 = 16/4 = 4

Hence r = a3/a2 = a2/a1 = 4 and a1 = 1,

a9 = a1*r⁸

a9 = 1*(4)⁸  

= (4)⁸

(ii) Given a2/a1 = (1/2)/(-3/4) = -2/3

a3/a2 = (-1/3)/(1/2) = -2/3

Hence, r = a3/a2 = a2/a1 = -2/3 and a1 = -3/4

a10 = a1*r⁹

a10 = (-3/4)*(-2/3)⁹

= (1/2)(2/3)⁸

(iii) Given a2/a1 = 0.06/0.3 = 0.2

a3/a2 = 0.012/0.06 = 0.2

Hence r = a3/a2 = a2/a1 = 0.2 and a1 = 0.3

a8 = a1*r⁷

a8 = 0.3*(0.2)⁷

(iv) Given a2/a1 = (ax)/(1/(a³x³)) = a⁴x⁴

a3/a2 = (a⁵x⁵)/(ax) = a⁴x⁴

Hence r = a3/a2 = a2/a1= a4x4 and a1 =1/(a³x³)

a12 = a1*r¹¹

a8 = (1/(a³x³))*((a⁴x⁴)¹¹)

     = a⁴¹x⁴¹

(v) Given a2/a1 = (1/√3)/√3 = 1/3

a3/a2 = (1/3√3)/(1/√3) = 1/3

Hence r = a3/a2 = a2/a1= 1/3 and a1 = √3

an = a1*r^(n-1)

     = √3*(1/3)^(n-1)   

问题4.从GP 2 / 27、2 / 9、2 / 3，………，162的末尾找到第四个项。

解决方案：

Given a2/a1 = (2/9)/(2/27) = 3

a3/a2 = (2/3)/(2/9) = 3

Hence r = a3/a2 = a2/a1= 3 and a1 = 2/27

After reversing the GP it becomes 162, ……….,, 2/3, 2/9, 2/27.

a1’=162 and r’= 1/r =1/3

a4’= a1′ * r’³

     = 162 * (1/3)³

     = 6.

问题5.进度的哪个项0.004,0.02,0.1，…是12.5?

解决方案：

Given a2/a1= 0.02/0.004= 5

a3/a2 = 0.1/0.02 = 5

Hence, r=a3/a2 = a2/a1= 5 and a1=0.004

nth term of sequence is given by

an = a1*(r)^n-1

an=(0.004)*(5)^n-1

Let the nth term be 12.5.

Hence, (0.004)*(5)^n-1=12.5

(5)^n-1= 12.5/0.004

(5)^n-1= 3125

(5)^n-1= (5)⁵

Hence, n-1=5

n=5+1= 6

问题6. GP的哪个任期：

(i)√2，1 /√2，1 /2√2，1 /4√2，…………..是1 /(512√2)吗?

(ii)2、2√2、4，………………..是128吗?

(iii)√3，3，3√3，……………………。是729吗?

(iv)1 / 3、1 / 9、1 / 27，…………是1/19683吗?

解决方案：

(i) In given sequence a1=√2, a2= 1/√2

Common ratio(r) = a2/a1 = (1/√2)/√2 =1/2 and a1 = √2

Let the nth term be 1/(512√2)

a1*r^n-1 = 1/(512√2)

√2*(1/2)^n-1 = 1/(512√2)

(1/2)^n-1 = 1/1024

(1/2)^n-1 = (1/2)10

Hence, n-1= 10

n=10+1 =11.

(ii) In given sequence a1=2 ,a2= 2√2

Common ratio(r)=a2/a1= 2√2/2 =  √2 and a1=2

Let the nth term be  1/(512√2)

a1*r^n-1 = 128

2*(√2)^n-1= 128

(√2)^n-1=64

(√2)^n-1= (√2)12

Hence n-1= 12

n= 12+1 =13.

(iii) In given sequence a1= √3, a2= 3

Common ratio(r)=a2/a1 = 3/√3 = √3 and a1 = √3

Let the nth term be  729

a1*r^n-1 = 729

√3*(√3)^n-1= 729

(√3)ⁿ = 729

(√3)ⁿ = (√3)¹¹

Hence n-1= 11

n= 11+1 =12.

(iv) In given sequence a1= 1/3, a2= 1/9

Common ratio(r)=a2/a1 = 1/3 and a1 = 1/3

Let the nth term be  1/19683

a1*r^n-1 = 1/19683

1/3*(1/3)^n-1 = 1/19683

(1/3)ⁿ = 1/19683

(1/3)ⁿ = (1/3)⁸

Hence n = 8.

问题7.哪个进展18，-12、8，…………..是512/729?

解决方案：

In given sequence a1=18 ,a2 =-12

Common ratio(r)= -12/18 = -2/3  and a1=18

Let the nth term be  512/729 

a1*r^n-1 = 512/729

(18)*(-2/3)^n-1 = 512/729 

(-2/3^)n-1 = 256/6561

(-2/3)^n-1 = (-2/3)⁸

Hence n-1 = 8

n = 8+1 = 9

问题8.从GP 1/2 1 / 6、1 / 18、1 / 54，…………..1 / 4373的末尾找到第四个项?

解决方案：

In given sequence a1 = 1/2, a2 = 1/6

Common ratio(r)= (1/6)/(1/2) =1/3 and a1 = 1/2

After reversing the GP it becomes 1/4374,……..1/54, 1/18, 1/6, 1/2

Common ratio(r’)= 3 and a1 = 1/4374

4th term from end of original GP becomes 4th from beginning of reversed GP

Hence required term is given by

(a1′)*(r’^4-1) = (a1′)*(r’)³         

                  = (1/4374)*(3)³

                  = (27/4374)

                  = 1/162

问题9. GP的第四学期是27，第七学期是729，找到GP吗?

解决方案：

Let a be first term of required GP and r be common ratio.

Given, a4= 27 and a7 = 729

(a1*r⁶) / (a1*r³) = 729/27 

r³ = 729/27 = 27

r³ = 3³

r=3

Now, put r=3 in  a1*r³=27

a1*(3)³ = 27 

a1 = 1

Thus, the given GP is 1,3, 9, 27,……….

问题10. GP的第7个学期是第4个学期的8倍，第5个学期是48个学期。

解决方案：

Let a be first term of required GP and r be common ratio.

Given, a7 = 8*a4 and a5 = 48

a7/a4 = 8

(a1*r⁶)/(a1*r³) = 8

r³ = 8

r=2

Now put r = 2 in a5 = 48

a1*r⁴ = 48

a1* (2)⁴ = 48

a1* 16 = 48 

a1 = 3

Thus, the given GP is 3, 6, 12, 24,…………..