第10类RD Sharma解决方案–第9章算术级数–练习9.4 |套装2

问题14. AP的第四项是第一项的三倍，而第七项是第三项的两倍乘以1。求出第一项和共同的区别。

解决方案：

Let’s assume that the first term and the common difference of the A.P to be a and d respectively.

Given that,

4th term of an A.P. is three times the first

a₄ = 3(a)

As we know that, to find nth term in an A.P = a + (n – 1)d

a + (4 – 1)d = 3a

3d = 2a ⇒ a = 3d/2 ———(i)

and,

7th term exceeds twice the third term by 1

a₇ = 2(a3) + 1

a + (7 – 1)d = 2(a + (3–1)d) + 1

a + 6d = 2a + 4d + 1

a – 2d +1 = 0 ——–(ii)

Using (i) in (ii), we get

3d/2 – 2d + 1 = 0

3d – 4d + 2 = 0

d = 2

therefore, putting d = 2 in (i), we get a

a = 3

Hence, the first term is 3 and the common difference is 2.

问题15：找到一个AP的第二项和第n项，其第六项为12，第八项为22。

解决方案：

Given that,

a₆ = 12 and a₈ = 22

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore,

a₆ = a + (6-1)d = a + 5d = 12 ——–(i)

and,

a₈ = a + (8-1)d = a + 7d = 22 ———(ii)

Solving (i) and (ii), we get

(ii) – (i)

a + 7d – (a + 5d) = 22 – 12

2d = 10

d = 5

now put d in (i) we get,

a + 5(5) = 12

a = 12 – 25

a = -13

Hence for the A.P: a = -13 and d = 5

so, the nth term is given by an = a + (n-1)d

an = -13 + (n-1)5 = -13 + 5n – 5

an = 5n – 18

Hence, the second term is given by a₂ = 5(2) – 18 = 10 – 18 = -8

问题16.两位数字可以被3整除吗?

解决方案：

The first 2 digit number divisible by 3 is 12 and the last 2 digit number divisible by 3 is 99.

this forms an A.P,

12, 15, 18, 21, …., 99

Where, a = 12 and d = 3

now find the number of terms in this A.P

99 = 12 + (n-1)3

99 = 12 + 3n – 3

90 = 3n

n = 90/3 = 30

Hence, there are 30 two digit numbers divisible by 3.

问题17. AP由60个词组成。如果第一项和最后一项分别为7和125，则找到第32个项。

解决方案：

Given that,

An A.P of 60 terms

and a = 7 and a60 = 125

As we know that, to find nth term in an A.P = a + (n – 1)d

a₆₀ = 7 + (60 – 1)d = 125

7 + 59d = 125

59d = 118

d = 2

Therefore, the 32nd term is given by

a₃₂ = 7 + (32 -1)2 = 7 + 62 = 69

Hence a₃₂ = 69

问题18. AP的第四项和第八项之和是24，第六和第十项的总和是34。找出AP的第一项和共同点

解决方案：

Given that,

The sum of 4th and 8th terms of an A.P. is 24 and

the sum of the 6th and 10th terms is 34

a₄ + a₈ = 24

and, As we know that, to find nth term in an A.P = a + (n – 1)d

[a + (4-1)d] + [a + (8-1)d] = 24

2a + 10d = 24

a + 5d = 12 ———(i)

a6 + a10 = 34

[a + 5d] + [a + 9d] = 34

2a + 14d = 34

a + 7d = 17 ———-(ii)

Subtracting (i) form (ii), we get

a + 7d – (a + 5d) = 17 – 12

2d = 5

d = 5/2

Using d in (i) we get,

a + 5(5/2) = 12

a = 12 – 25/2

a = -1/2

Hence, the first term is -1/2 and the common difference is 5/2.

问题19. AP的第一项为5，而其第100项为-292。查找此AP的第50个术语

解决方案：

Given that,

a = 5 and a100 = -292

As we know that, to find nth term in an A.P = a + (n – 1)d

a100 = 5 + 99d = -292

99d = -297

d = -3

Hence, the 50th term is

a₅₀ = a + 49d = 5 + 49(-3) = 5 – 147 = -142

问题20：找到AP的a30 – a20

(i)-9，-14，-19，-24(ii)a，a + d，a + 2d，a + 3d，……

解决方案：

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore, a₃₀ – a₂₀ = (a + 29d) – (a + 19) =10d

(i) Given A.P. -9, -14, -19, -24

Here, a = -9 and d = -14 – (-9) = = -14 + 9 = -5

therefore, a30 – a20 = 10d

= 10(-5) = -50

(ii) Given A.P. a, a+d, a+2d, a+3d, ……

therefore, a₃₀ – a₂₀ = (a + 29d) – (a + 19d) =10d

问题21：为AP a，a + d，a + 2d，……写表达式an – ak。

(i)第11个学期为5，第13个学期为79。

(ii) ₁₀ – ₅ = 200

(iii)第20个学期比第18个学期多10个学期。

解决方案：

Given A.P. a, a+d, a+2d, …..

therefore, an = a + (n-1)d = a + nd –d

And, ak = a + (k-1)d = a + kd – d

an – ak = (a + nd – d) – (a + kd – d)

= (n – k)d

(i) Given that 11th term is 5 and 13th term is 79,

where n = 13 and k = 11,

a₁₃ – a₁₁ = (13 – 11)d = 2d

79 – 5 = 2d

d = 74/2 = 37

(ii) Given that, a₁₀ – a₅ = 200

(10 – 5)d = 200

5d = 200

d = 40

(iii) Given that, 20th term is 10 more than the 18th term.

a₂₀ – a₁₈ = 10

(20 – 18)d = 10

2d = 10

d = 5

问题22：确定x的给定值是否为给定AP的第n个项

(i)25、50、75、100 ，； x = 1000(ii)-1，-3，-5，-7，…； x = -151

(iii)5½，11、16½，22等。。。 x = 550(iv)1、21 / 11、31 / 11、41 / 11，…； x = 171/11

解决方案：

(i) Given that, A.P. 25, 50, 75, 100, ……,1000

where, a = 25 d = 50 – 25 = 25

nth term= 1000

As we know that, to find nth term in an A.P = a + (n – 1)d

1000 = 25 + (n-1)25

1000 = 25 + 25n – 25

n = 1000/25

n = 40

(ii) Given that, A.P. -1, -3, -5, -7, …., -151

where, a = -1 d = -3 – (-1) = -2

nth term = -151

As we know that, to find nth term in an A.P = a + (n – 1)d

-151 = -1 + (n-1)(-2)

-151 = -1 – 2n + 2

n = 152/2

n = 76

(iii) Given that, A.P. 5½, 11, 16½, 22, …, 550

where, a = 5½ d = 11 – (5½) = 5½ = 11/2

nth term = 550

As we know that, to find nth term in an A.P = a + (n – 1)d

550 = 5½ + (n-1)(11/2)

550 x 2 = 11+ 11n – 11

1100 = 11n

n = 100

(iv) Given that, A.P. 1, 21/11, 31/11, 41/11, 171/11

where, a = 1 d = 21/11 – 1 = 10/11

nth term = 171/11

As we know that, to find nth term in an A.P = a + (n – 1)d

171/11 = 1 + (n-1)10/11

171 = 11 + 10n – 10

n = 170/10

n = 17

问题23. AP的第八学期是其第二学期的一半，而第十一学期则超过其第四学期的三分之一乘以1.找到第15学期。

解决方案：

Given that,

a₈ = 1/2(a₂)

a₁₁ = 1/3(a₄) + 1

As we know that, to find nth term in an A.P = a + (n – 1)d

a₈ = 1/2(a₂)

a + 7d = 1/2(a + d)

2a + 14d = a + d

a + 13d = 0 ——-(i)

And, a₁₁ = 1/3(a4) + 1

a + 10d = 1/3(a + 3d) + 1

3a + 30d = a + 3d + 3

2a + 27d = 3 ——–(ii)

By solving (i) and (ii), by (ii) – 2x(i)

2a + 27d – 2(a + 13d) = 3 – 0

d = 3

Putting d in (i) we get,

a + 13(3) = 0

a = -39

Hence, the 15th term a₁₅ = -39 + 14(3) = -39 + 42 = 3

问题24.找到第三项为16，第七项比第五项超出12的算术级数。

解决方案：

Given that,

a₃ = 16 and a₇ = a₅ + 12

As we know that, to find nth term in an A.P = a + (n – 1)d

a + 2d = 16…… (i)

and,

a + 6d = a + 4d + 12

2d = 12

d = 6

Using d in (i), we get

a + 2(6) = 16

a = 16 – 12 = 4

Hence, the A.P is 4, 10, 16, 22, …….

问题25. AP的第7项是32，而其13项是62。找到AP

解决方案：

Given that,

a₇ = 32 and a₁₃ = 62

From a_n – a_k = (a + nd – d) – (a + kd – d) = (n – k)d

a₁₃ – a₇ = (13 – 7)d = 62 – 32 = 30

6d = 30

d = 5

Now,

a₇ = a + (7 – 1)5 = 32

a + 30 = 32

a = 2

Hence, the A.P is 2, 7, 12, 17, ……

问题26. AP 3、10、17的哪个任期。会比第13个学期多84个吗?

解决方案：

Given that,

A.P. 3, 10, 17, ….

where, a = 3 and d = 10 – 3 = 7

a_n = a₁₃ + 84 (Given)

As we know that, to find nth term in an A.P = a + (n – 1)d

3 + (n – 1)7 = 3 + (13 – 1)7 + 84

3 + 7n – 7 = 3 + 84 + 84

7n = 168 + 7

n = 175/7

n = 25

Hence, the 25th term which is 84 more than its 13th term.

问题27.两个算术级数具有相同的共同点。他们的第100个条款之间的差是100，他们的第1000个条款之间有什么区别?

解决方案：

Let’s assume that the two A.Ps be A.P1 and A.P2

For A.P1 the first term is a and the common difference is d

and for A.P2 the first term is b and the common difference is d

Given that,

a100 – b100 = 100

(a + 99d) – (b + 99d) = 100

a – b = 100

Now, the difference between their 1000th terms is,

(a + 999d) – (b + 999d) = a – b = 100

Hence, the difference between their 1000th terms is also 100.