10类NCERT解决方案–第5章算术级数–练习5.3 |套装2

问题11.如果AP的前n个项之和为4n-n ² ，则第一个项(即S ₁ )是什么?前两个项的总和是多少?第二学期是什么?类似地，找到第3，第10和第n个项。

解决方案：

Given:

S_n = 4n−n²

First term can be obtained by putting n=1, 

a = S₁ = 4(1) − (1)² = 4−1 = 3

Also,

Sum of first two terms = S₂= 4(2)−(2)² = 8−4 = 4

Second term, a₂ = S2 − S1 = 4−3 = 1

Common difference, d = a₂−a₁ = 1−3 = −2

Also,

Nth term, a_n = a+(n−1)d

= 3+(n −1)(−2)

= 3−2n +2

= 5−2n

Therefore, a₃ = 5−2(3) = 5-6 = −1

a₁₀ = 5−2(10) = 5−20 = −15

Therefore, the sum of first two terms is equivalent to 4. The second term is 1.

And, the 3rd, the 10th, and the nth terms are −1, −15, and 5 − 2n respectively.

问题12。找到被40除以的前40个正整数的总和。

解决方案：

The first positive integers that are divisible by 6 are 6, 12, 18, 24 ….

Noticing this series, 

First term, a = 6 

Common difference, d= 6.

Sum of n terms, we know,

S_n = n/2 [2a +(n – 1)d]

Substituting the values, we get

S₄₀ = 40/2 [2(6)+(40-1)6]

= 20[12+(39)(6)]

= 20(12+234)

= 20×246

= 4920

问题13.找到8的前15倍的总和。

解决方案：

The first few multiples of 8 are 8, 16, 24, 32…

Noticing this series, 

First term, a = 8 

Common difference, d = 8.

Sum of n terms, we know,

S_n = n/2 [2a+(n-1)d]

Substituting the values, we get,

S₁₅ = 15/2 [2(8) + (15-1)8]

= 15/2[6 +(14)(8)]

= 15/2[16 +112]

= 15(128)/2

= 15 × 64

= 960

问题14.查找0到50之间的奇数之和。

解决方案：

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.

Therefore, we can see that these odd numbers are in the form of A.P.

Now,

First term, a = 1

Common difference, d = 2

Last term, l = 49

Last term is equivalent to, 

l = a+(n−1) d

49 = 1+(n−1)2

48 = 2(n − 1)

n − 1 = 24

Solving for n, we get,

n = 25

Sum of nth term,

S_n = n/2(a +l)

Substituting these values, 

S₂₅ = 25/2 (1+49)

= 25(50)/2

=(25)(25)

= 625

问题15.一份建筑工程合同规定了对延迟超过一定日期的罚款，具体规定如下：卢比。第一天200卢比。第二天250卢比。第三天以此类推，罚款300卢比。比前一天多了50。如果承包商将工作推迟了30天，承包商必须支付多少钱作为罚款。

解决方案：

The given penalties form and A.P. having first term, a = 200 and common difference, d = 50.

By the given constraints, 

Penalty that has to be paid if contractor has delayed the work by 30 days = S₃₀

Sum of nth term, we know,

S_n = n/2[2a+(n -1)d]

Calculating, we get, 

S₃₀= 30/2[2(200)+(30 – 1)50]

= 15[400+1450]

= 15(1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty for 30 days delay.

问题16.一笔700卢比的款项将用于为学校学生的整体学习表现提供七笔现金奖励。如果每个奖赏比之前的奖赏少20卢比，请找出每个奖赏的价值。

解决方案：

Let us assume the cost of 1st prize be Rs. P.

Then, cost of 2nd prize = Rs. P − 20

Also, cost of 3rd prize = Rs. P − 40

These prizes form an A.P., with common difference, d = −20 and first term, a = P.

Given that, S₇ = 700

Sum of nth term, 

S_n = n/2 [2a + (n – 1)d]

Substituting these values, we get,

7/2 [2a + (7 – 1)d] = 700

Solving, we get,

a + 3(−20) = 100

a −60 = 100

a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

问题17：在学校里，学生们想到在学校内外种植树木以减少空气污染。决定每个班级的每个部分将种植的树木数量与他们正在研究的班级的树木数量相同，例如，第一类的一部分将种植一棵树，第二类的一部分将会种植一棵树。种植2棵树，依此类推，直到第十二类。每个课程分为三个部分。学生将种植几棵树?

解决方案：

The number of trees planted by the students form an AP, 1, 2, 3, 4, 5………………..12

Now,

First term, a = 1

Common difference, d = 2−1 = 1

Sum of nth term,

S_n = n/2 [2a +(n-1)d]

S₁₂ = 12/2 [2(1)+(12-1)(1)]

= 6(2+11)

= 6(13)

= 78

Number of trees planted by 1 section of the classes = 78

Therefore,

Number of trees planted by 3 sections of the classes = 3×78 = 234

问题18：一个螺旋形由连续的半圆组成，中心在A和B处交替发生，如图所示，半径为A，半径分别为0.5、1.0 cm，1.5 cm，2.0 cm，…………。这样的由13个连续的半圆组成的螺旋的总长度是多少? (取π= 22/7)

解决方案：

We know,

Perimeter of a semi-circle = πr

Calculating ,

P₁ = π(0.5) = π/2 cm

P₂ = π(1) = π cm

P₃ = π(1.5) = 3π/2 cm

Where, P₁, P₂, P₃ are the lengths of the semi-circles respectively.

Now, this forms a series, such that,

π/2, π, 3π/2, 2π, ….

P₁ = π/2 cm

P₂ = π cm

Common difference, d = P₂ – P₁ = π – π/2 = π/2

First term = P₁= a = π/2 cm

Sum of nth term, 

Sn = n/2 [2a + (n – 1)d]

Therefore, Sum of the length of 13 consecutive circles is;

S₁₃ = 13/2 [2(π/2) + (13 – 1)π/2]

Solving, we get,

=  13/2 [π + 6π]

=13/2 (7π)

= 13/2 × 7 × 22/7

= 143 cm

问题19：200根原木以下列方式堆叠：下一行20根原木，下一行19根，下一行19根，依此类推。 200条日志放在几行中，第一行放在几行中?

[1]

解决方案：

The numbers of logs in rows are in the form of an A.P. 20, 19, 18…

Given, 

First term, a = 20 and common difference, d = a2−a1 = 19−20 = −1

Let us assume a total of 200 logs to be placed in n rows.

Thus, S_n = 200

Sum of nth term,

Sn = n/2 [2a +(n -1)d]

S12 = 12/2 [2(20)+(n -1)(-1)]

400 = n (40−n+1)

400 = n (41-n)

400 = 41n−n²

Solving the eq, we get,

n²−41n + 400 = 0

n²−16n−25n+400 = 0

n(n −16)−25(n −16) = 0

(n −16)(n −25) = 0

Now,

Either (n −16) = 0 or n−25 = 0

n = 16 or n = 25

By the nth term formula,

a_n = a+(n−1)d

a₁₆ = 20+(16−1)(−1)

= 20−15

= 5

And, the 25th term is, 

a₂₅ = 20+(25−1)(−1)

= 20−24

= −4

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5, since the number of logs can’t be negative as in case of 25th term.

问题20.在马铃薯比赛中，将一个铲斗放置在起点处，该铲斗距第一个马铃薯5 m，而其他马铃薯则以直线间隔3 m放置。排队有十个土豆。

一名竞争对手从水桶开始，捡起最近的马铃薯，随即向后跑，将其丢入水桶，然后跑回以捡起下一个马铃薯，跑向水桶以将其放入，然后她以相同的方式继续前进，直到所有的土豆都在桶里。竞争对手要跑的总距离是多少?

[提示：要拿起第一个马铃薯和第二个马铃薯，竞争对手跑的总距离(以米为单位)为2×5 + 2×(5 + 3)]

解决方案：

The distances of potatoes from the bucket are 5, 8, 11, 14…, which form an AP.

Now, we know that  the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.

Therefore, distances to be run w.r.t distances of potatoes is equivalent to, 

10, 16, 22, 28, 34,……….

We get, a = 10 and d = 16−10 = 6

Sum of nth term, we get,

S₁₀ = 12/2 [2(20)+(n -1)(-1)]

= 5[20+54]

= 5(74)

Solving we get, 

= 370

Therefore, the competitor will run a total distance of 370 m.