第 11 课 RD Sharma 解决方案 - 第 19 章算术级数 - 练习 19.2 |设置 2

General Formula

Tn = a + (n – 1) * d

where,

Tn is the nth term

a is the first term

d = common difference

   

Nth term form end

Tn = l – (n – 1) * d

where,

Tn is the nth term

l is the first term

d = common difference

问题 13. 如果 AP 的第 (m + 1) 项是第 ( ⁿ + 1) 项的两倍，证明^第(3m + 1) 项是^第(m+ ⁿ + 1) 项的两倍。

解决方案：

We are given, a_m+1 = 2 a_n+1

To prove, a_3m+1 = 2a_m+n+1

Lets suppose we have first term as a and common difference as d, then we are given

a + md = 2(a + nd)

a + md = 2a + 2nd

a = md- 2nd —–1

Now LHS = a_3m+1

= a + 3md

= md – 2nd +3md putting value of a

= 4md – 2nd

Now RHS = 2a_m+n+1

= 2( a + (m+n)d)

= 2( a +md +nd)

= 2 (md-2nd+md+nd) putting value of a

= 2(2md – nd)

= 4md – 2nd

So we can see that LHS = RHS.

问题 14. 如果 AP 9, 7, 5, ... 的第ⁿ项与 AP 15, 12, 9, ... 的^第n 项相同 求 N。

解决方案：

We are given two AP. For first AP

a₁ = 9

d₁ = 7 – 9 = -2

T_1n = a₁ + (n – 1)d₁

= 9 + (n – 1)(-2)

= 11 – 2n

For second AP,

a₂ = 15

d₂ = 12 – 15 = -3

T_2n = a₂ + (n – 1)d₂

= 15 +(n-1)(-3)

= 18 – 3n

According to question,

T_1n = T_2n

11 – 2n = 18 – 3n

n = 7

Hence 7^th term is same in both AP.

问题 15. i) 找出以下算术级数的第 12 项：3, 5, 7, 9,... 201。

解决方案：

We are given, first term, a = 3

last term, l = 201

common difference, d = 5 – 3 = 2

n = 12

So, the 12th term from last is , = 201 – (12 -1)*2

= 201 – 22

= 179

Hence, the 12th term from the end is 179.

ii) 找出以下算术级数末尾的第 12 项：3、8、13、... 253。

解决方案：

We are given, first term, a = 3

last term, l = 253

common difference, d = 8 – 3 = 5

n = 12

So, the 12th term from last is, = 253 – (12 -1)*5

= 253 – 55

= 198

Hence, the 12th term from the end is 198.

iii) 找出以下算术级数末尾的第 12 项：1, 4, 7, 10,… 88。

解决方案：

We are given, first term, a = 1

Last term, l = 88

common difference, d = 4 – 1 = 3

n = 12

So, the 12th term from last is, = 88 – (12 -1)*3

= 88 – 33

= 55

Hence, the 12th term from the end is 55.

问题 16. AP 的第 4 项是第 1 项的 3 倍，第 7 项比第 3 项高 1 倍。求第一项和共同差。

解决方案：

Let’s suppose first term is a and common difference is d

We are given, a₄ = 3a₁

a₇ = 2a₃ +1

So ,

a + 3d = 3a

2a – 3d = 0 —–1

And a + 6d = 2(a+2d) +1

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

a -2d = -1 —–2

On solving 1 and 2

2a – 3d = 0

-2a + 4d = 2

d = 2

Putting value of d in equation 1

a = (3*d)/2

a = 6/2 = 3

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

问题 17. 找出第 6 项为 12 且第 8 项为 22 的 AP 的第 2 项和第 N 项。

解决方案：

Let’s suppose first term is a and common difference is d

We are given, a₆ = 12

a₈ = 22

So, a + 5d =12 —–1

a + 7d = 22 —–2

On solving 1 and 2

a + 7d – (a + 5d )= 22 – 12

7d – 5d = 10

d = 5

Putting value of d in equation 1

a + 25 = 12

a = 12- 25

a = -13

So second term is, a + d = -13 + 5 = -8

And n^th term is T_n = -13 + (n-1)*5

= 5n – 18

问题 18. 有多少个两位数可以被 3 整除？

解决方案：

First two digit number that is divisible bye 3 is 12. So a =3

Next two digit number that is divisible bye 3 is 15. So common difference is 15 – 12 = 3

Last two digit number that is divisible bye 3 is 99.

So we can apply general formula,

3 + (n-1)*3 = 99

(n-1)*3 = 96

n -1 = 32

n =33

Hence, 33 two digit numbers are divisible by 3.

问题 19. AP 由 60 个术语组成。如果第一项和最后一项分别为 7 和 125，则查找第 32 项。

解决方案：

We are given, a = 7

l = 125

total terms = 60

If we have total 60 terms, then last term would be a₆₀

a + 59d =125

Putting value of a,

7 + 59d = 125

59d = 118

d= 2

So, 32nd term a₃₂ = a + 31d

= 7 + 31*2

= 7 + 62

= 69

Hence, 32nd term is 69.

问题 20. AP 的第 4 项和第 8 项之和为 24，第 6 项和第 10 项之和为 34。求 AP 的第一项和公差

解决方案：

Let’s suppose first term is a and common difference is d. We are given,

a₄ + a₈ = 24

a₆ + a₁₀ = 34

a+ 3d + a +7d = 24

2a + 10d = 24

a + 5d = 12 —–1

a + 5d + a+9d = 34

2a + 14d = 34

a + 7d = 17 ——2

On solving 1 & 2 we get,

a + 7d – (a + 5d) = 17 – 12

a + 7d – a – 5d = 5

2d = 5

d = 5/2

Putting value of d in equation 1

a + 5*(5/2) = 12

a = 12 – 25/2

a = (24- 25) /2

a = -1/2

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

问题 21. 1 到 1000 之间有多少个数除以 7 余数为 4？

解决方案：

Let’s make an AP of those numbers which when divided by 7 leaves remainder 4.

First number which when divided by 7 leaves remainder 4 is 4. So a₁ = 4

Next number which when divided by 7 leaves remainder 4 is 11. So a₂ = 11

Largest 3 digit number which when divided by 7 leaves remainder 4 is 998. So a_n = 998

So our AP is 4, 11, …. 998 and common difference is 11-4 = 7

So,

4 + (n – 1)*7 = 998

(n – 1)*7 = 994

n – 1 = 142

n =143

Hence, 143 numbers are which when divided by 7 leave remainder 4.

问题 22. AP 的第一项和最后一项分别是 a 和 l。证明开头第 n 项和结尾第 n 项之和为 a + l。

解决方案：

Let’s suppose common difference is d.

We know that nth term from beginning is a + (n-1)d

And nth term from end is l – (n – 1)*d

To prove: nth term from beginning + nth term from end = a + l

LHS= nth term from beginning + nth term from end

= a + (n-1)d + l – (n – 1)*d

= a +l

= RHS

Hence proved.

问题 23. 如果一个 AP 是这样的₄ /a ₇ = 2/3，找到一个₆ /a ₈ 。

解决方案：

Let’s suppose first term is a and common difference is d.

We are given, (a + 3d)/(a+6d) = 2/3

3(a + 3d) =2(a+6d)

3a + 9d = 2a + 12d

a = 3d

Now, we have to find a₆/a₈

= (a + 5d)/ (a+7d)

Putting value of a

= (3d + 5d)/(3a + 7d)

= 8d/10d

= 4/5

Hence a₆/a₈ is 4/5.