第10类RD Sharma解决方案–第9章算术级数–练习9.2

问题1：证明由_n = 5n – 7定义的序列是AP，找到其共同的区别。

解决方案：

Given:

 a_n = 5n – 7

Now putting n = 1, 2, 3, 4,5 we get,

a₁ = 5.1 – 7 = 5 – 7 = -2

a₂ = 5.2 – 7 = 10 – 7 = 3

a₃ = 5.3 – 7 = 15 – 7 = 8

a₄ = 5.4 – 7 = 20 – 7 = 13

We can see that,

a₂ – a₁ = 3 – (-2) = 5

a₃ – a₂ = 8 – (3) = 5

a₄ – a₃ = 13 – (8) = 5

Since, the successive difference of list is same i.e 5

∴ The given sequence is in A.P and have common difference of 5

问题2：证明由_n = 3n ² – 5定义的序列不是AP

解决方案：

Given: 

a_n = 3n² – 5

Now putting n = 1, 2, 3, 4 we get,

a₁ = 3.1.1 – 5= 3 – 5 = -2

a₂ = 3.2.2 – 5 = 12 – 5 = 7

a₃ = 3.3.3 – 5 = 27 – 5 = 22

a₄ = 3.4.4 – 5 = 48 – 5 = 43

We can see that,

a₂ – a₁ = 7 – (-2) = 9

a₃ – a₂ = 22 – 7 = 15

a₄ – a₅ = 43 – 22 = 21

Since, the successive difference of list is not the same 

∴ The given sequence is not in A.P 

问题3：序列的一般术语由_n = -4n + 15给出。该序列是AP吗?如果是这样，找到它的第15项和共同的区别。

解决方案：

Given:

 a_n = -4n + 15

Now putting n = 1, 2, 3, 4 we get,

a₁ = -4.(1) + 15 = -4 + 15 = 11

a₂ = -4.(2) + 15 = -8 + 15 = 7

a₃ = -4.(3) + 15 = -12 + 15 = 3

a₄ = -4.(4) + 15 = -16 + 15 = -1

We can see that,

a₂ – a₁ = 7 – (11) = -4

a₃ – a₂ = 3 – 7 = -4

a₄ – a₃ = -1 – 3 = -4

Since, the successive difference of list is same i.e -4

∴ The given sequence is in A.P and have common difference of -4

Hence, the 15th term will be

a₁₅ = -4(15) + 15 = -60 + 15 = -45

And, a₁₅ = -45

问题4：用第n个项写出序列：

(i) _n = 3 + 4n

解决方案：

Given:

a_n = 3 + 4n

Now putting n = 1, 2, 3, 4 we get,

a₁ = 3 + 4.1 = 7 

a₂ = 3 + 4.2 = 11

a₃ = 3 + 4.3 = 15

a₄ = 3 + 4.4 = 19

∴ The sequence is 7, 11, 15, 19

We can see that,

a₂ – a₁ = 11 – (7) = 4

a₃ – a₂ = 15 – (11) = 4

a₄ – a₃ = 19 – (15) = 4

Since, the successive difference of list is same i.e 4

∴ The given sequence is in A.P

(ii) _n = 5 + 2n

解决方案：

Now putting n = 1, 2, 3, 4 we get,

a₁ = 5 + 2.1 = 7

a₂ = 5 + 2.2 = 9

a₃ = 5 + 2.3 = 11

a₄ = 5 + 2.4 = 13

∴ The sequence is 7, 9, 11, 13

We can see that,

a₂ – a₁ = 9 – (7) = 2

a₃ – a₂ = 11 – (9) = 2

a₄ – a₃ = 13 – (11) = 2

Since, the successive difference of list is same i.e 2

∴ The given sequence is in A.P

(iii)a _n = 6 – n

解决方案：

Now putting n = 1, 2, 3, 4 we get,

a₁ = 6 – 1 = 5

a₂ = 6 – 2 = 4

a₃ = 6 – 3 = 3

a₄ = 6 – 4 = 2

∴ The sequence is 5, 4, 3, 2

We can see that,

a₂ – a₁ = 4 – (5) = -1

a₃ – a₂ = 3 – (4) = -1

a₄ – a₃ = 2 – (3) = -1

Since, the successive difference of list is same i.e -1

∴ The given sequence is in A.P

(iv)a _n = 9 – 5n

解决方案：

Now putting n = 1, 2, 3, 4 we get,

a₁ = 9 – 5.1 = 4

a₂ = 9 – 5.2 = -1

a₃ = 9 – 5.3 = -6

a₄ = 9 – 5.4 = -11

∴ The sequence is 4, -1, -6, -11

We can see that,

a₂ – a₁ = -1 – (4) = -5

a₃ – a₂ = -6 – (-1) = -5

a₄ – a₃ = -11 – (-6) = -5

Since, the successive difference of list is same i.e -5

∴ The given sequence is in A.P

问题5：AP的第n个项是6n +2。找到共同的不同之处。

解决方案：

Now putting n = 1, 2, 3, 4 we get,

a₁ = 6.1 + 2 = 8

a₂ = 6.2 + 2 = 14

a₃ = 6.3 + 2 = 20

a₄ = 6.4 + 2 = 26

We can see that,

a₂ – a₁ = 14 – (8) = 6

a₃ – a₂ = 20 – (14) = 6

a₄ – a₃ = 26 – (20) = 6

Hence, the common difference is 6

问题6：证明说n个序列为AP的序列是否正确

(i)a _n = 2n – 1

解决方案：

Now putting n = 1, 2, 3, 4 we get,

a₁ = 2.1 – 1 = 1

a₂ = 2.2 – 1 = 3

a₃ = 2.3 – 1 = 5

a₄ = 2.4  – 1 = 7

We can see that,

a₂ – a₁ = 3 – (1) = 2

a₃ – a₂ = 5 – (3) = 2

a₄ – a₃ = 7 – (5) = 2

Since, the successive difference of list is same i.e 2

Hence, the given sequence is in A.P

(ii) _n =3n²+ 5

解决方案：

Now putting n = 1, 2, 3, 4 we get,

a₁ = 3.1.1 + 5 = 8

a₂ = 3.2.2 + 5 = 17

a₃ = 3.3.3 + 5 = 32

a₄ = 3.4.4 + 5 = 53

We can see that,

a₂ – a₁ = 17 – (8) = 9

a₃ – a₂ = 32 – (17) = 15

a₄ – a₃ = 53 – (32) = 21

Since, the successive difference of list is not the same 

Hence, the given sequence is not in A.P

(iii)a _n = 1 + n +n²

解决方案：

Now putting n = 1, 2, 3 we get,

a₁ = 1 + 1 + 1.1 = 3

a₂ = 1 + 2 + 2.2 = 7

a₃ = 1 + 3 + 3.3 = 13

We can see that,

a₂ – a₁ = 7 – (3) = 4

a₃ – a₂ = 13 – (7) = 6

Since, the successive difference of list is not the same

Hence, the given sequence is not in A.P