第 11 课 RD Sharma 解决方案 - 第 19 章算术级数 - 练习 19.7 |设置 1

问题 1. 一个男人节省了卢比。 10年内16500。在第一次之后的每一年，他都节省了卢比。比他在退去的一年中所做的多了 100 个。第一年他存了多少钱？ ²

解决方案：

Let the money saved by man in the first year be Rs. x then,

as given in the problem, his saving increases by Rs. 100 per year so,

x + (x + 100)+ (x + 200)+ (x + 300)+ ….. + (x + 900) = 16500

⇒ 10x + 100 (1+2+3+….+9) = 16500

⇒ 10x + 100*(9 * 10)/2 = 16500 (Since, sum of n natural numbers = n(n+1)/2)

⇒ 10x + 4500 = 16500

⇒ 10x = 12000

⇒ x = 1200

Therefore, the man saved Rs. 1200 in the first month.

问题 2。一个男人节省了卢比。第一年 32 卢比。第二年 36 元，这样他的储蓄增加了卢比。每年4个。找出他的储蓄将在什么时候达到卢比。 200。

解决方案：

Let the time taken by the man to save Rs. 200 be n years.

Here, 32 + 36 + 40 +…. = 200

We can notice, first term, a=32, common difference, d=4 and sum, S_n = 200.

We know, Sn = n [2a + (n-1)*d]/2 using this formula, we get

⇒ 200 = n [2*32 + (n-1)*4]/2

⇒ 400 = 64n + 4n² -4n

⇒ 4n² + 60n – 400 = 0

⇒ n² + 15n -100 = 0

⇒ n -5n + 20n -100 = 0

⇒ n(n-5) + 20(n-5) = 0

⇒ (n-5) (n+20) = 0

⇒ n=5 or n= -20

We consider, n =5 since years cannot be in negative.

Therefore, the man saves Rs. 200 in 5 years.

问题 3. 一名男子安排偿还 Rs 的债务。 3600 40 年分期付款，形成一个算术系列。当 30 期还清时，他死了，剩下三分之一的债务未偿，求第一期的价值。

解决方案：

According to the question, 40 annual installments are paid in form an arithmetic series, so

Using the sum formula, Sn = n [2a + (n-1)*d]/2

⇒ 3600 = 40[2a + (40-1)d]/2

⇒ 2a + 39d = 180 —————————- (eqn.1)

Also, sum of first 30 installments is two-third of 3600 which is 2400

⇒ 30[2a + (30-1)d] = 2400

⇒ 2a + 29d = 160 —————————- (eqn.2)

Subtracting eqn.2 from eqn.1, we get

10d = 20 ⇒d = 2

Putting the value of d in eqn.1 we get a= 51

Therefore, the value of first installment is Rs. 51

问题 4. 一家收音机制造商第三年生产了 600 台，第七年生产了 700 台。假设产品每年以固定数量均匀增长，求 (i) 第一年的产量，(ii) 七年的总产品和 (iii) 第 10 年的产品。

解决方案：

According to question, a₃ = a + 2d = 600 and a₇ = a + 6d = 700

Solving for a and d, we get a = 550 and d = 25.

Therefore, (i) total 550 radios were produced in the first year.

(ii) Total radios produced in first seven years

= 7 [550 + 700]/2 [since, S_n = n (a + l) /2]

= 2375

(iii) Product produced in 10th year = a + 9d = 550 + 9×25 = 775

问题 5. 有 25 棵树在 5 米的距离与一口井成一条直线。井距最近的树的距离为 10 米。园丁从井开始分别给所有的树浇水，给每棵树浇水后，他回到井里为下一棵树取水。找出园丁浇灌所有树木所需的总距离。

解决方案：

As given in the question, the distance of the well from the nearest tree = 10 m. and in total there are 25 trees separated by a distance of 5 m. all in a line with the well.

The total distance covered by the gardener in watering 25 trees will be given as:

2 [10 + 15 + 20 + 25 + ….. + 135] (Multiplying it by 2 because he moves back also)

⇒ 2 * [25 * [2*10 + (25-1)5]/2]

⇒ total distance = 3500 meters.

Hence, the total distance the gardener will cover in order to water all the trees is 3500 meters.

问题 6. 一个人被雇来数卢比。 10710 卢比。每分钟180个半小时。在此之后，他以卢比计算。每分钟比前一分钟少3。找出他计算全部金额所花费的时间。

解决方案：

As given in the question. for half an hour he counts at the rate of Rs. 180 every minute, so amount = 180 * 30 = 5400. Now the amount left = 10710 – 5400 = 5310

Then, he counts Rs. 3 less every minute than the preceding minute, so

(180 -3) + (180 – 2×3) + (180 – 3×3) + …. = 5310

Let the time taken to count this amount be n years, so

5310 = n [(180-3) + (n-1) (-3)]/2

⇒ 5310 = n [200 – 3n]

⇒ 3n² -200n +5310 = 0

⇒ n = 59

Therefore, total time taken to count the entire amount of Rs. 10710 is 59+30 = 89 minutes.

问题 7. 一台设备的成本是某个工厂卢比。 600,000。如果它贬值，第一年贬值 15%，第二年贬值 13.5%，第三年贬值 12%，依此类推，10 年末贬值是多少，所有百分比都适用于原始成本。

解决方案：

The piece deprecates by 15% of its value in first year which is 15% of 600,000 = 90,000

Therefore, value of the piece after first year = 600,000 – 90,000 = 510,000.

Value deprecates by 13.5% of its value in second year i.e, 13.5% of 600,000 = 81,000

Value deprecates by 12% of its value in second year i.e, 12% of 600,000 = 72,0006–

Therefore, total deprecation in 10 years, can be calculated as:

S = 10 *[2* 81000 + (10-1) (-9000)] /2

= 5 * 81000 = 405000

Therefore, the cost of piece after 10 years will be = 600,000 – 405,000 = 1050000

问题 8. 一位农民以卢比购买了一台二手拖拉机。 12000. 他支付卢比。 6000 现金，并同意每年分期支付余额。 500 加上未付金额的 12% 利息。拖拉机花了他多少钱。

解决方案：

Total cost of the tractor

= 6000 + [(500 + 12% of 6000) + (500 + 12% of 5500) + (500 + 12% of 5000) ….. to 12 times]

= 6000 +6000 +12% of (6000 + 5500 + 5000 + …. + 12 times)

= 12000 + 12% of [12 (6000 + 500)/2]

= 12000 + 12% of [6 * 6500]

= 12000 + 4680

= 16680

Therefore, in total cost of the tractor to the farmer is Rs. 16680

问题 9. Shamshad Ali 以卢比购买了一辆踏板车。 22000. 他支付卢比。 4000 现金，并同意每年分期支付余额。 1000 加上未付金额的 10% 利息。踏板车花了他多少钱。

解决方案：

Total cost of the scooter

= 4000 + [(1000 + 10% of 18000) + (1000 + 10% of 17000) + (1000 + 10% of 16000) ….. to 18 times]

= 4000 +18000 +10% of (1800+ 1700 + 1600 + …. + 18 times)

= 22000 + 10% of [18 (1800 + 100)/2]

= 22000 + 10% of [9 * 19000]

= 22000 + 17100

= 39100

Therefore, in total cost of the tractor to the farmer is Rs. 39100

问题 10。一个人的收入是卢比。第一年300,000，他收到了卢比的增加。在接下来的 19 年中，他每年的收入为 10000。求总金额，他在20年内收到。

解决方案：

First year income of person, a = 300,000

here, common difference, d = 10,000

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

⇒ Sn = 20 [2*300000 + (20 – 1) * 10000]/2

⇒ Sn =10 * [600000 + 190000]

⇒ Sn = 7900000

Therefore, total amount, he received in 20 years is Rs. 7900000