10类NCERT解决方案-第5章算术级数-练习5.4

问题1. AP的哪个任期：121、117、113，…。 。 。，是第一个否定词吗?

[提示：找到n表示<0]

解决方案：

Given the AP series is 121, 117, 113, . . .,

Here, first term, a = 121 and Common difference, d = 117-121= -4

n^th term formula, a_n = a+(n −1)d

Hence,

a_n = 121+(n−1)(-4)

= 121-4n+4

=125-4n

To find the first negative term of the series, an < 0 Therefore,

125-4n < 0

125 < 4n

n>125/4

n>31.25

Therefore, the first negative term of the series is 32^nd term (n=32).

which will be, a₃₂ =125-4(32) = -3

问题2. AP的第三项和第七项之和为6，其乘积为8。找到AP的前十六项之和。

解决方案：

n^th term formula, a_n = a+(n −1)d

Third term, a₃ = a+(3 -1)d

a₃ = a + 2d

And Seventh term, a₇= a+(7-1)d

a₇ = a + 6d

According to given conditions, 

a₃ + a₇ = 6 …………………………….(i)

a₃ × a₇ = 8 ……………………………..(ii)

Substituting the values in eqn. (i), we get,

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

a = 3–4d …………………………………(iii)

Now substituting the values in eqn. (ii), we get,

(a+2d)×(a+6d) = 8…………………..(iv)

Putting the value of a from eqn. (iii) in eqn. (iv), we get,

(3–4d +2d)×(3–4d+6d) = 8

(3 –2d)×(3+2d) = 8

3² – (2d)² = 8         (using the identity, (a+b)(a-b) = a²-b²)

9 – 4d² = 8

4d² = 9-8 = 1

d = √(1/4)

d = ±1/2

d = 1/2 or -1/2

So now, if

d = 1/2,

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1

and if d = -1/2,

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5

Sum of n^th term of AP is: S_n = n/2 [2a +(n – 1)d]

So, when a = 1 and d=1/2

Then, the sum of first 16 terms are;

S₁₆ = 16/2 [2 +(16-1)1/2] = 8(2+15/2) 

S₁₆ = 76

And when a = 5 and d= -1/2

Then, the sum of first 16 terms are;

S₁₆ = 16/2 [2.5+(16-1)(-1/2)] = 8(5/2) 

S₁₆ = 20

问题3.梯子的横档相距25厘米。 (见图)。梯级的长度从底部的45厘米均匀减小到顶部的25厘米长度。如果顶部和底部梯级之间的距离为2½m，则梯级所需的木材长度是多少?

[提示：横档数= 250/25 +1]

解决方案：

As it is given that, 

Distance between the rungs of the ladder is 25 cm.

Distance between the top rung and bottom rung of the ladder will be (in cm) = 2 ½ × 100 = 250 cm

Hence, the total number of rungs = 250/25 + 1 = 11

As we can observe here, that, the ladder has rungs in decreasing order from top to bottom. Thus, the rungs are decreasing in an order of AP.

So, According to given condition 

First term, a = 45

Last term, l = 25

Number of terms, n = 11

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

Now, as we know, sum of n^th terms is equal to, S_n= (n/2)(a+ l)

S_n= 11/2(45+25) = 11/2(70) = 385 cm

Hence, the length of the wood required for the rungs is 385cm.

问题4.一排房屋的编号从1到49连续编号。证明存在x的值，使得在编号为x的房屋之前的房屋数量的总和等于房屋数量的总和。跟着它。找出x的值。

[提示：S _{x – 1} = S ₄₉ – S _x ]

解决方案：

According to the given statement, 

Row houses are numbered in a row that are in the form of AP, which is as follows: 

1

2

3

…..

…..

….

49

Here, according to the given condition, 

First term, a = 1

Common difference, d=1

and last term = l

Let us say the number of x^th houses can be represented as: S_{x – 1} = S₄₉ – S_x

Where S_x represents to sum of x^th houses.

(Value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it)

As, Sum of n^th term of AP, S_n= n/2[a+l]

Sum of number of houses beyond x house = S_x-1

S_x-1 = (x-1)/2[1+(x-1)]

S_x-1 = x(x-1)/2 ………………………………………(i)

By the given condition, we can write,

S₄₉ – S_x = {49/2[1+(49)}–{x/2(1+x)}

= 25(49) – x(x + 1)/2 ………………………………….(ii)

As eqn. (i) and (ii) are equal, So

x(x-1)/2 = 25(49) – x(x-1)/2

x(x-1) = 25(49)

x = ±35

As we know, the number of houses cannot be a negative number. 

Hence, the value of x is 35.

问题5.足球场上的一个小露台包括15个台阶，每个台阶长50 m，由坚固的混凝土建造。每个台阶的上升高度为1/4 m，胎面的高度为1/2 m(见图)。计算建造露台所需的混凝土总量。

[提示：第一步所需的混凝土体积= 1/4×1/2×50 m ³ ]

解决方案：

As we can see from the given figure, 

1^st step is ¼ m high, ½ m wide and 50 m long

2^nd step is (¼+¼ = ½ m) high, ½ m wide and 50 m long and, 

3^rd step is (3×¼ = 3/4 m) high, ½ m wide and 50 m long. 

and so on

Hence, we can conclude that the height of step increases by ¼ m each time when width and length is ½ m and 50 m respectively. 

So, the height of steps forms a series AP in such a way that;

¼, ½ , ¾, 1, 5/4, ……..

Volume of steps = Volume of Cuboids

= Length × Breadth × Height

Now,

Volume of concrete required to build the first step = ¼ ×½ ×50 = 25/4

Volume of concrete required to build the second step =½ ×½×50 = 25/2

Volume of concrete required to build the second step = ¾ × ½ ×50 = 75/2

Volume of steps = ½ × 50 × (¼ + 2/4 + 3/4 + 4/4 + 5/4 + …….) …………………….(1)

Now, we can see the height of concrete required to build the steps, are in AP series;

Thus, applying the AP series concept to the height,

First term, a = 1/4

Common difference, d = 2/4-1/4 = 1/4

n = 15

As, the sum of n terms is : S_n = n/2[2a+(n-1)d] 

S_n = 15/2(2×(1/4 )+(15 -1)1/4)

S_n = 15/2 (4)

S_n = 30

Hence, up solving eqn. (1), we get

Volume of steps = ½ × 50 × S_n

= ½ × 50 × 30

= 750 m³

Hence, the total volume of concrete required to build the terrace is 750 m³.