第 11 类 RD Sharma 解决方案 - 第 19 章算术级数 - 练习 19.7 |设置 2

问题 11。一个男人开始偿还贷款作为第一期卢比。 100. 如果他将分期付款增加卢比。 5 每个月，第 30 期他会付多少钱？

解决方案：

In the first installment, the man pays Rs. 100, so a₁ = 100 and in next installment the man pays Rs. 105, so a₂ = 105.

So we can conclude, common difference, d = 105 – 100 = 5

Thus, the money he will pay in 30th installment, a₃₀ = a₁ + 29d = 100 + 29 x 5 = 100 + 145 = 245

Hence, he will pay Rs. 245 in the 30th installment.

问题 12. 聘请木匠建造 192 个窗框。第一天他做了五个框架，之后的每一天他都比前一天多做了两个框架。他花了多少时间才能完成这项工作？

解决方案：

On the first day, the carpenter makes 5 frames, so a = 5 and thereafter each day he makes 2 extra frames, so d = 2.

Total frames to be made, Sn = 192

We know the formula,

$S_n = \frac{n[2a + (n-1)d]}{2}$

$\Rightarrow 192 = \frac{n[2\times 5 + (n-1)2]}{2}$

$\Rightarrow n^2 + 4n -192 = 0$

$\Rightarrow n^2 + 16n -12n -192 = 0$

$\Rightarrow (n+16)(n-12) = 0$

$\Rightarrow n = -16, 12$

We will neglect n= -16 and consider n = 12

Therefore, the carpenter took 12 days to make 192 frames.

问题 13. 我们知道三角形的内角和是 180°。证明具有 3,4,5,6…的多边形的内角之和。形成算术级数。求 21 边多边形的内角和。

解决方案：

The sum of interior angles of a polygon having n-sides is = (n-2) x 180°

Thus, for a polygon with 3 sides, we have a₃ = (3-2) x 180° = 180°

similarly, for a polygon with 4 sides, we have a₄ = (4-2) x 180° = 360°

for a triangle with 5 sides, we have a₅ = (5-2) x 180° = 540°

Now, a₄ – a₃ = 360° – 180° = 180°

also, a₅ – a₄ = 540° – 360° = 180°, we are getting the same common difference for every succeeding and preceding term, hence we can say that it forms an AP.

Now,  the sum of the interior angles for a 21 sided polygon = (21-2) x 180° = 3420°

问题 14. 在马铃薯比赛中，20 个马铃薯以 4 m 的间隔排成一排，第一个马铃薯距离起点 24 m。参赛者需要将土豆一次一个地带回起点。他会跑多远才能把所有的土豆都带回来？

解决方案：

Since, there are in total 20 potatoes and they are placed in a line at intervals of 4 m, therefore. n = 20 and d=4  

First potato is placed at a distance of 24 m, so a₁ = 24, now a₂ = 28, similarly a₂₀ = 24+ 19 x 4 = 100.

We know the formula, $S_n = \frac{n}{2}\left [ a+l \right ]$  

therefore, $S_n = \frac{20}{2}\left [ 24 + 100 \right ] = 1240$

The contestant has to bring back the potatoes hence, he will run back and forth, so total distance covered = 1240 x 2  =2480 m.

问题 15. 一名男子接受了一个起薪为卢比的职位。每月5200。据了解，他将收到自动增加的卢比。下个月和之后的每个月都有 320 个。

(i) 找出他第十个月的薪水。

(ii) 第一年的总收入是多少？

解决方案：

Given: initial salary, a = Rs. 5200 and d = Rs. 320.

(i) his salary in 10th month = a + 9d = 5200 + 9 x 320 = 8080

(ii)  his total earnings during the first year

 $= \frac{12}{2}\left [ 2 \times 5200 + (12-1)\times 320 \right ]$

$6\left [ 10400 + 3520 \right ] = 6 \times 13920 = 83520$

问题 16。一个男人节省了卢比。 20年66000。在第一年之后的每一年，他都节省了卢比。比他去年存的钱多了200。第一年他存了多少钱？

解决方案：

Given: Sn = 66000, n=20, d=200.

We know the formula,

$S_n = \frac{n[2a + (n-1)d]}{2}$

$\Rightarrow 66000 = \frac{20[2\times a + (20-1)200]}{2}$

$\Rightarrow 66000 = 10[2\times a + 19\times 200]$

$\Rightarrow 6600 -3800 = 2a$

$\Rightarrow a = 1400$

The man saved Rs. 1400 in the first year.

问题 17。在板球队锦标赛中，有 16 支球队参加。他们之间将颁发 8000 卢比作为奖金。如果最后一名的队伍获得 275 卢比的奖金，并且连续完成名次的奖金增加相同，那么第一名的队伍将获得多少奖金？

解决方案：

Total number of teams participating in the tournament, n = 16

Total prize money, Sn = 8000

The last placed team received prize money, a = 275

Suppose, for every successive team the prize money increases by d, then by the sum formula

$\Rightarrow 8000 = \frac{16[2\times 275 + (16-1)d ]}{2}$

$\Rightarrow 1000 = [550 + 15d ]$

$\Rightarrow d = 30$

therefore, the amount received by the first placed team = a+15d = 275 + 15 x 30 = 725

Hence, the first placed team received Rs 725 as the prize money.