第10类RD Sharma解决方案–第9章算术级数–练习9.3

问题1：对于以下算术级数，写出第一项a和共同的差d：

(i)-5，-1、3、7，…………

解决方案：

Given sequence is -5, -1, 3, 7, …………

∴ First term is -5

And, common difference = a₂ – a₁ = -1 – (-5) = 4

∴ Common difference is 4

(ii)1 / 5、3 / 5、5 / 5、7 / 5，……

解决方案：

Given sequence is 1/5, 3/5, 5/5, 7/5, ……

∴ First term is 1/5

And, common difference = a₂ – a₁ = 3/5 – (1/5) = 2/5

∴ Common difference is 2/5

(iii)0.3、0.55、0.80、1.05，…………

解决方案：

Given sequence is 0.3, 0.55, 0.80, 1.05, …………

∴ First term is 0.3

And, common difference = a₂ – a₁ = 0.55 – (0.3) = 0.25

∴ Common difference is 0.25

(iv)-1.1，-3.1，-5.1，-7.1，…………..

解决方案 ：

Given sequence is -1.1, -3.1, -5.1, -7.1, …………..

∴ First term is -1.1

And, common difference = a₂ – a₁ = -3.1 – (-1.1) = -2.0

∴ Common difference is -2.0

问题2：当第一项a和共同差d如下时，写出算术级数：

(i)a = 4，d = -3

解决方案：

Given:

a₁ = 4 and d= -3

Therefore, arithmetic progression is: a, a + d, a + 2d, a + 3d, ……

⇒ 4, 4 – 3, 4 + 2(-3), 4 + 3(-3), ……

⇒ 4, 1, – 2, – 5, – 8 ……..

∴ A.P will be 4, 1, – 2, – 5, – 8 ……..

(ii)a = -1，d = 1/2

解决方案：

Given:

a₁ = -1, d= 1/2

Therefore, arithmetic progression is: a, a + d, a + 2d, a + 3d, ……

⇒ -1, -1 + 1/2, -1, 2½, -1 + 3½, …

⇒ -1, -1/2, 0, 1/2

∴ A.P will be -1, -1/2, 0, 1/2

(iii)a = -1.5，d = -0.5

解决方案：

Given:

a₁ = -1.5, d = -0.5

Therefore, arithmetic progression is: a, a + d, a + 2d, a + 3d, ……

⇒ -1.5, -1.5, -0.5, –1.5 + 2(– 0.5), –1.5 + 3(– 0.5)

⇒ – 1.5, – 2, – 2.5, – 3, …….

∴ A.P will be – 1.5, – 2, – 2.5, – 3, …….

问题3：在以下哪种情况下，形成的数字序列将形成AP?

(i)挖一口井的第一米费用为₹ 150，随后每米增加₹20。

解决方案：

Given:

Cost of digging a well for the first metre = ₹150

Therefore,

Cost for the second metre = ₹150 + ₹20 = ₹170

Cost for the third metre = ₹170 + ₹20 = ₹190

Cost for the fourth metre = ₹190 + ₹20 = ₹210

Since, 20 is added fo each succeeding metre

Hence, the sequence will be (In rupees) and is A.P 

150, 170, 190, 210, ………..

∴ The given sequence is in A.P and common difference is 20

(ii)当真空泵每次移除气缸中剩余气体的1/4时，气缸中存在的空气量。

解决方案：

Air removed for the first time = (1 x 1/4) = 1/4

Remaining air = 1 – 1/4 = 3/4

Air removed for the second time = (3/4 x 1/4) = 3/16

Remaining air = 3/4 – 3/16 = 9/16

Air removed for the third time = (9/16 x 1/4) = 9/64

Remaining air = 9/16 – 9/64 = 27/64

∴The sequence will be 1, 3/4, 9/16, 27/64

Here, a₂ – a₁ = 3/4 – (1) = -1/4

a₃ – a₂ = 9/16 – (3/4) = -3/16

Since, the successive difference of list is not same

∴ The given sequence is not in A.P

(iii)Divya按年利率10％的复合利率存入1000卢比。第一年末，第二年末，第三年末等的金额。

解决方案：

Given:

Divya deposited Rs 1000 at compound interest of 10% p.a

So, the amount at the end of first year is = 1000 + 0.1(1000) = Rs 1100

And, the amount at the end of second year is = 1100 + 0.1(1100) = Rs 1210

And, the amount at the end of third year is = 1210 + 0.1(1210) = Rs 1331

Here, a₂ – a₁ = 1210 – 1100 = 110

a₃ – a₂ = 1331 – 1210 = 121

Since, the successive difference of list is not same

∴ The given sequence is not in A.P

问题4：找到共同的区别，并为以下每个算术级数写下四个项：

(i)1，-2，-5，-8，……..

解决方案：

Given:

First term = 1, Second term = -2

Common difference = -2 – (1) = -3

Now,

Fifth term = -8 + (-3) = -11

Sixth term = -11 + (-3) = -14

Seventh term = -14 + (-3) = -17

Eighth term = -17 + (-3) = -20

(ii)0，-3，-6，-9，……

解决方案：

Given:

First term = 0, Second term = -3

Common difference = -3 – (0) = -3

Now,

Fifth term = -9 + (-3) = -12

Sixth term = -12 + (-3) = -15

Seventh term = -15 + (-3) = -18

Eighth term = -18 + (-3) = -21

(iii)-1，1/4，3/2，……..

解决方案：

Given:

First term = -1, Second term = 1/4

Common difference = 1/4 – (-1) = 5/4

Now,

Fifth term = 3/2 + (5/4) = 11/4

Sixth term = 11/4 + (5/4) = 4

Seventh term = 4 + (5/4) = 21/4

Eighth term = 21/4 + (5/4) = 26/4

(iv)-1，– 5/6，– 2/3，…………..

解决方案：

Given:

First term = -1, Second term = -5/6

Common difference = -5/6 – (-1) = 1/6

Now,

Fifth term = -2/3 + (1/6) = -1/2

Sixth term = -1/2 + (1/6) = -1/3

Seventh term = -1/3 + (1/6) = -1/6

Eighth term = -1/6 + (1/6) = 0

问题5：证明无论实数a和b是多少，第n^个项a + nb的序列始终是AP。有什么共同点?

解决方案：

a_n = a + nb

Let n= 1, 2, 3, 4, 5, ……….

a₁ = a + b

a₂ = a + 2b

a₃ = a + 3b

a₄ = a + 4b

a₅ = a + 5b

Now,

a₂ – a₁ = (a + 2b) – (a + b) = b

a₃ – a₂ = (a + 3b ) – (a + 2b) = b

a₄ – a₁ = (a + 4b) – (a + 3b) = b

a₅ – a₄ = (a + 5b) – (a + 4b) = b

Hence, proved

问题6：找出以下哪些序列是算术级数。对于那些算术级数，找出它们的共同区别

(i)3、6、12、24，………………………………

解决方案：

Now,

a₂  – a₁ = 6 – (3) = 3

a₃ – a₁ = 12 – (6) = 6

a₄ – a₁ = 24 – (12) = 12

Since, the successive difference of list is not same

∴ The given sequence is  not in A.P 

(ii)0，-4，-8，-12，………………………………………………

解决方案：

Now,

a₂  – a₁ = -4 – (0) = -4

a₃ – a₂ = -8 – (-4) = -4

a₄ – a₃ = -12 – (-8) = -4

Since, the successive difference of list is same

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

(iii)1 / 2、1 / 4、1 / 6、1 / 8

解决方案：

Now,

a₂  – a₁ = 1/4 – (1/2) = -1/4

a₃ – a₂ = 1/6 – (1/4) = -1/12

a₄ – a₃ = 1/8 – (1/6) = -1/24

Since, the successive difference of list is not same

∴ The given sequence is not in A.P

(iv)12、2，-8，-18，…………………………..

解决方案：

Now,

a₂  – a₁ = 2 – (12) = -10

a₃ – a₂ = -8 – (2) = -10

a₄ – a₃ = -18 – (-8) = -10

Since, the successive difference of list is same

∴ The given sequence is in A.P

(v)3、3、3、3，…………………………………………

解决方案：

Now,

a₂  – a₁ = 3 – (3) = 0

a₃ – a₂ = 3 – (3) = 0

a₄ – a₃ = 3 – (3) = 0

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 0

(vi)p，p + 90，p + 180，………………..

解决方案：

Now,

a₂  – a₁ = p+90 – (p) = 90

a₃ – a₂ = p+180 – (p+90) = 90

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 90

(vii)1.0、1.7、2.4，…………………………。

解决方案：

Now,

a₂  – a₁ = 1.7 – (1.0) = 0.7

a₃ – a₂ = 2.4 – (1.7) = 0.7

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 0.7

(viii)-225，-425，-625，………………………………。

解决方案：

Now,

a₂  – a₁ = -425 – (-225) = -200

a₃ – a₂ = -625 – (-425) = -200

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is -200

(IX)10，10 + 2 ^5，10 + 2 ^6，10 + 2 ^7， ……………………………………..

解决方案：

Now,

a₂  – a₁ = 10+2⁵ – (10) = 32

a₃ – a₂ = 10+2⁶ – (10+2⁵) = 32

a₄ – a₃ = 10+2⁷ – (10+2⁶) = 64

Since, the successive difference of list is not same

∴ The given sequence is in A.P

(x)a + b，(a + 1)+ b，(a + 1)+(b + 1)，(a + 2)+(b + 1)，……………………………… …。

解决方案：

Now,

a₂  – a₁ = ((a+1) + b) – (a + b) = 1

a₃ – a₂ = ((a+1)+(b+1)) – ((a+1)+b) = 1

a₄ – a₃ = ((a+2)+(b+1)) – ((a+1)+(b+1)) = 1

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 1

(XI)1 ^{2，3 2，5 2，7 2，………………………………………………………}

解决方案：

Now,

a₂  – a₁ = (3² – 1²) = 8

a₃ – a₂ = (5² – 3²) = 16

a₄ – a₃ = (7² – 5²) = 24

Since, the successive difference of list is not same

∴ The given sequence is not in A.P

(XII)1 ^{2，5 2，7 2，73，..} …………………………………………………………

解决方案：

Now,

a₂  – a₁ = 5² – 1² = 24

a₃ – a₂ = 7² – 5² = 24

a₄ – a₃ = 73 – 7² = 24

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 24

问题7：找出AP的共同点，并写下以下两个术语：

(i)51、59、67、75，……。

解决方案：

Given: 

First term = 51, Second term = 59

Common difference = 59 – 51 = 8

Now,

Fifth term = 75 + 8 = 83

Sixth term = 83 + 8 = 91

(ii)75、67、59、51，…………

解决方案：

Given:

 First term = 75, Second term = 67

Common difference = 67 – 75 = -8

Now,

Fifth term = 51 – 8 = 43

Sixth term = 43 – 8 = 35

(iii)1.8、2.0、2.2、2.4，……。

解决方案：

Given:

 First term = 1.8, Second term = 2.0

Common difference = 2.0 – 1.8 = 0.2

Now,

Fifth term = 2.4 + (0.2) = 2.6

Sixth term = 2.6 + (0.2) = 2.8

(iv)0、1 / 4、1 / 2、3 / 4，…………..

解决方案：

Given: 

First term = 0, Second term = 1/4

Common difference = 1/4 – 0 = 1/4

Now,

Fifth term = 3/4 + (1/4) = 1

Sixth term = 1 + (1/4) = 5/4

(v)119，136，153，170，…………..

解决方案：

Given: 

First term = 119, Second term = 136

Common difference = 136 – 119 = 17

Now,

Fifth term = 170 + (17) = 187

Sixth term = 187 + (17) = 204