10类NCERT解决方案-第5章算术级数-练习5.1

问题1.在以下哪种情况下，所涉及的数字列表会进行算术级数运算，为什么?

(一世)。票价为Rs时，每公里后计程车费。前15公里和Rs为15。每增加1公里增加8个。

解决方案：

Initially fare = 0

After completing 1 km, fare = 15

After completing 2 km, fare = 15 + 8 = 23

After completing 3 km, fare = 23 + 8 = 31

And so on

So you can write fare in series as

15, 23, 31, 39,……… and so on

Here you can see that the first term is 15 and the difference between any two-term is 8. So, this series is in arithmetic progression.

Note: A series is said to be in arithmetic progression if any term can be found out by adding a fixed number to the previous term. The first term is denoted as and the fixed number is known as a difference which is denoted by d.

In the above series a = 15 and d = 8

(ii)。当真空泵一次清除气缸中剩余空气的1/4时，气缸中存在的空气量。

解决方案：

Initial volume of air in cylinder = πR²H

Amount of air removed by vacuum pump = πR²H/4

Remaining air = 3 * πR²H/4

Again amount of air removed by vacuum pump = 3 * πR²H/16

Remaining air = 3 * πR²H/4 – 3 * πR²H/16 = 9 * πR²H/16 

and so on

The series can be written as: πR²H, 3 * πR²H/4, 9*πR²H/16,……… and so on

You can see here is no common difference between any two-term. So, this series is not arithmetic progression.

(iii)。每米挖掘一口井的成本为Rs。第一米150卢比。随后的每米50。

解决方案：

Cost of digging well for first meter is 150 rupees

Cost of digging well for two meters is 150 + 50 = 200 rupees

Cost of digging well for three meters is 200 + 50 = 250 rupees

And so on

The series for cost of digging a well look like 150, 200, 250, 300,……… and so on

Here you can see that after the first term, any term can be found out by adding 50 to the previous term. Hence, the above series is in arithmetic progression with the first term 150 and the common difference is 50.

(iv)。每年的帐户金额为10000卢比，复利利率为8％。

解决方案：

Formula for compound interest is given by P(1 + r/100)ⁿ where P is the principal amount, r is the rate of interest and n is the time duration.

So, the series can be written as: 10000(1 + 8/100), 10000(1 + 8/100)²,10000(1 + 8/100)³,……… and so on

As time duration increases, compound interest will increase exponentially. Hence, this can’t form a series in arithmetic series because there is no fixed common difference between any two terms. 

问题2.当第一项a和共同差d如下时，写出AP的四个项：

(一世)。 a = 10，d = 10

解决方案：

Given:

a₁ = a = 10

d = 10

Now we find the remaining terms:

Second term(a₂) = a₁ + d = 10 + 10 = 20

Third term(a₃) = a₂ + d = 20 + 10 = 30

Fourth term(a₄) = a₃ + d = 30 + 10 = 40

and so on.

So, four terms of this A.P. will be as follows

10, 20, 30, 40,….

(ii)a = -2，d = 0

解决方案：

Given:

a₁ = a = -2

d = 0

Now we find the remaining terms:

Second term(a₂) = a₁ + d = -2 + 0 = -2

Third term(a₃) = a₂ + d = -2 + 0 = -2

Fourth term(a₄) = a₃ + d = -2 + 0 = -2

and so on.

So, four terms of this A.P. will be as follows

-2, -2, -2, -2,….

(iii)。 a = 4，d = -3

解决方案：

Given:

a₁ = a = 4

d = -3

Now we find the remaining terms:

Second term(a₂) = a₁ + d = 4 + (-3) = 1

Third term(a₃) = a₂ + d = 1 + (-3) = -2

Fourth term(a₄) = a₃ + d = -2 + (-3) = -5

and so on.

So, four terms of this A.P. will be as follows

4, 1, -2, -5,….

(iv)。 a = -1，d = 1/2

解决方案：

Given:

a₁ = a = -1

d = 1/2

Now we find the remaining terms:

Second term(a₂) = a₁ + d = -1 + 1/2 = -1/2

Third term(a₃) = a₂ + d = (-1/2) + (1/2) = 0

Fourth term(a₄) = a₃ + d = 0 + (1/2) = 1/2

and so on.

So, four terms of this A.P. will be as follows

-1, -1/2, 0, 1/2,….

(v)。 a = -1.25，d = -0.25

解决方案：

Given:

a₁ = a = -1.25

d = -0.25

Now we find the remaining terms:

Second term(a₂) = a₁ + d = -1.25 – 0.25 = – 1.50

Third term(a₃) = a₂ + d = -1.50 – 0.25 = -1.75

Fourth term(a₄) = a₃ + d = -1.75 – 0.25 = -2.00

and so on.

So, four terms of this A.P. will be as follows

-1.25, -1.50, -1.75, -2.00,….

问题3：对于以下AP，请写出第一项和共同点。

(一世)。 3，1，-1，-3，…。

解决方案：

From the above A.P., first term (a) = 3

Common difference (d) = second term – first term

= 1 – 3 = -2

(ii)。 -5，-1、3、7…。

解决方案：

From the above A.P., first term (a) = -5

Common difference (d) = second term – first term

= -1 – (-5)

= -1 + 5 = 4

(iii)。 1 / 3、5 / 3、9 / 3、13 / 3…。

解决方案：

From the above A.P., first term (a) = 1

Common difference (d) = second term – first term

= 5/3 – 1/3

= 4/3

(iv)。 0.6、1.7、2.8、3.9…。

解决方案：

From the above A.P., first term (a) = 0.6

Common difference (d) = second term – first term 

= 1.7 – 0.6

= 1.1

问题4.以下哪个是AP?如果它们形成AP，则找到共同的差异d并再写三个项。

(一世)。 2、4、8、16…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 4 – 2 = 2

d₂ = a₃ + a₂ = 8 – 4 = 4

So d₁ ≠ d₂ 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(ii)。 2、5 / 2、3、7 / 2等。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 5/2 – 2 = 1/2

d₂ = a₃ + a₂ = 3 – 5/2 = 1/2

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is 1/2.

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = 7/2 + 1/2 = 4

Sixth term(a₆) = a₅ + d = (4) + (1/2) = 9/2

Seventh term(a₇) = a₆ + d = 9/2 + (1/2) = 5

So, the next three terms are: 4, 9/2, 5

(iii)。 -1.2，-3.2，-5.2，-7.2等。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = -3.2 – (-1.2) = -2.0

d₂ = a₃ + a₂ = -5.2 – (-3.2) = -2.0

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is -2.0

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = (-7.2) + (-2.0) = -9.2

Sixth term(a₆) = a₅ + d = (-9.2) + (-2.0) = -11.2

Seventh term(a₇) = a₆ + d = (-11.2) + (-2.0) = -13.2

So, the next three terms are: -9.2, -11.2, -13.2

(iv)。 -10，-6，-2、2…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = -6 – (-10) = 4

d₂ = a₃ + a₂ = -2 – (-6) = 4

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is 4

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = 2 + 4 = 6

Sixth term(a₆) = a₅ + d = 6 + 4 = 10

Seventh term(a₇) = a₆ + d = 10 + 4 = 14

So, the next three terms are 6, 10, 14

(v)。 3，3 +√2，3 +2√2，3 +3√2，…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 3 + √2 – 3 = √2

d₂ = a₃ + a₂ = 3 + 2√2 – (3 + √2) = √2

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is √2

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = (3 + 3√2) + √2 = 3 + 4√2

Sixth term(a₆) = a₅ + d = (3 + 4√2) + √2 = 3 + 5√2

Seventh term(a₇) = a₆ + d = (3 + 5√2) + √2 = 3 + 6√2

So, the next three terms are 3+4√2, 3+5√2, 3+6√2

(vi)。 0.2、0.22、0.222、0.2222…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 0.22 – 0.2 = 0.02

d₂ = a₃ + a₂ = 0.222 – 0.22 = 0.002

So d₁ ≠ d₂ 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(vii)。 0，-4，-8，-12…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = -4 – 0 = -4

d₂ = a₃ + a₂ = -8 – (-4) = -4

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is -4

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = (-12) + (-4) = -16

Sixth term(a₆) = a₅ + d = (-16) + (-4) = -20

Seventh term(a₇) = a₆ + d = (-20) + (-4) = -24

So, the next three terms are -16, -20, -24

(viii)。 -1 / 2，-1 / 2，-1 / 2，-1 / 2…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = -1/2 – (-1/2) = 0 

d₂ = a₃ + a₂ = -1/2 – (-1/2) = 0 

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is 0

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = (-1/2) + 0 = -1/2

Sixth term(a₆) = a₅ + d = (-1/2) + 0 = -1/2

Seventh term(a₇) = a₆ + d = (-1/2) + 0 = -1/2

So, the next three terms are -1/2, -1/2, -1/2

(ix)。 1，3，9，27，…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 3 – 1 = 2

d₂ = a₃ + a₂ = 9 – 3 = 6

So d₁ ≠ d₂ 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(X)。 a，2a，3a，4a…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 2a – a = a

d₂ = a₃ + a₂ = 3a – 2a = a

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is a

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = 4a + a = 5a

Sixth term(a₆) = a₅ + d = 5a + a = 6a

Seventh term(a₇) = a₆ + d = 6a + a = 7a

So, the next three terms are 5a, 6a, 7a

(xi)。 a，a ² ，a ³ ，a ⁴ …。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = a² – a 

d₂ = a₃ + a₂ = a³ – a²  

So d₁ ≠ d₂ 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(xii)。 √2，√8，√18，√32，…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 2√2 – √2 = √2

d₂ = a₃ + a₂ = 3√2 – 2√2 = √2

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is √2

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = √32 + √2 = 5√2

Sixth term(a₆) = a₅ + d = 5√2 + √2 = 6√2

Seventh term(a₇) = a₆ + d = 6√2 + √2 = 7√2

So, the next three terms are 5√2, 6√2, 7√2

(xiii)。 √3，√6，√9，√12，…。

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = √6 – √3 = √3(√2 – 1)

d₂ = a₃ + a₂ = √9 – √6 = √3(√3 – √2)

So d₁ ≠ d₂ 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(十四)。 1 ^{2，3 2，5 2，7 2，…。}

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 9 – 1 = 8

d₂ = a₃ + a₂ = 25 – 9 = 16

So d₁ ≠ d₂ 

Hence, this series doesn’t form an AP because there is no fixed common difference.

(xv)。 1 ^{2，5 2，7 2，73，…。}

解决方案：

First we check the given series is AP or not by finding common difference:

d₁ = a₂ + a₁ = 25 – 1 = 24

d₂ = a₃ + a₂ = 49 – 25 = 24

So d₁ = d₂ 

Hence, the above series is in arithmetic progression and the common difference is 24

Now we find the remaining terms of the A.P.:

Fifth term(a₅) = a₄ + d = 73 + 24 = 97

Sixth term(a₆) = a₅ + d = 97 + 24 = 121

Seventh term(a₇) = a₆ + d = 121 + 24 = 145

So, the next three terms are 97, 121, 145