NCERT CHAPTER 5 Mathematics Solutions and explanation

 

Simple explanation and solution to the arithmetic progression 

What is an arithmetic progression? 

Observe these series:

Ex(1): 1, 2, 3, 4, 5, 6...

Ex(2): 2, 4, 6, 8, 10 ...

Ex(3): 1/2, 3/2, 5/2, 7/2...

In example 1, you can see the difference between two consecutive terms: 2-1 = 1, 3-2 = 1, 4-3 = 1... and so on. 

In example 2: 4-2 = 2

                      6-4 = 2

                      8-6 = 2

and so on. Similarly, the difference of the two terms in example 3 is 1/2.

A series in which consecutive terms differ by a constant (fixed) value is called arithmetic series. 

The first term is often represented by T1 or a, and the fixed difference is denoted by "d"

If we observe this type of series, we can write it as:

a, (a+d), (a+2d), (a+3d),...

The first term contains no d, or it may be written as (1-1)d

The second term contains one d and can be written as (2-1)d

The third term contains 2d and can be written as (3-1)d. 

....

In this manner, we can write the formula for the 10th term as a + (10-1)d.

So, in general, the nth term from the beginning can be written 

Tn = a + (n-1)d

If we write n = 1, 2, 3, 4...

We get the values of the first terms, T1, T2, T3, T4, and so on.

If we reverse the series and write from last to first, then:

Tn' = Tn +(n-1)(-d)

The negative sign is present because we have reversed the series. 

Tn' is the last term on reversing the series.

Example: 3, 5, 7, 9, and 11 is a given series. Common difference = 2

Reversing the series, we get 11, 9, 7, 5, 3; common difference (d) = -2

Solutions

1(i)  The taxi fare for the first km = 15 Rs

        Taxi fare after 1 Km = 8 Rs for each subsequent kilometers

T1 = 15 Rs, 

T2 = 15+8 = 23,

T3 = 23+8 = 31 Rs. ...

T2 - T1 = 8 Rs

T3-T2 = 8 Rs

Therefore, we can see that the differences between the terms are 8 = d

1(ii)  Let the initial volume of the cylinder in V = T1

Removal of the Gas for the first time = V/4

Remaining Volume = V - V/4 = 3V/4  = T2

Next removal = 1/4 * (3V/4)  = 3v/16

Remaining Volume = 3v/4 - (3V/16)

= 3v/4(1-1/4) = 3v/4 * 3/4

                                =(9/16) V

If we see, T3-T2 = 9/16V - 3/4V = -3V/16

T2-T1 = 3v/4 - v = -v/4

So, the difference is not the same.

It is not an A.P. 

1(iii) Cost of digging the well for the first meter = 150 Rs

         Cost for each subsequent meter =                     50 Rs

If T1 = 150 Rs, T2 = 150+50 = 200 Rs, T3 = 200+50 = 250 Rs

Here, T2-T1 = 50 , T3-T2 = 50

The common difference is constant, and hence it is an A.P.

1(iv)  Given P = 10000 Rs

r = 8% 

Amount after the 1 year = P(1 + r/100) for the first year

                                       = 10,000(1+8/100)

 Amount after the two years = 10,000(1+8/100)^2 

^ = read as "power" or "exponent"

Amount after 3 years = 10,000(1+8/100)^3 Rs

If we consider these amounts as T1, T2, and T3 

T2-T1 = 10,000(1+8/100)[(1+8/100-1)] = 10,000(1+8/100)(8/100)

T3-T2 = 10,000(1+8/100)^2 {1+8/100-1}

            = 10,000(1+8/100)^2 *(8/100) 

Since T2-T1 is not equal to T3-T2, it is not an A.P. 

Q2(i) 

  Given the first term and the common difference, a = 10, d = 10

T1 = 10

T2 = a+d = 10+10 = 20

T3 = a+2d = 10+2*10 = 30

T4 = a+3d = 10+10*3 = 40 

(ii) a =-2, d = 0

 Since a = 2 and d = 0, there will be no difference in the next terms. All the terms will remain same. 

T1 = -2, T2 = -2, T3 =-2, T4 =-2 

(iii) a = 4, d = -3

 T1 = 4

T2 = a+d = 4+(-3) = 4-3 = 1

T3 = a + 2d = 4 + 2 * (-3) = 4 - 6 = -2

T4 = a + 3d = 4 + 3 * (-3) = 4 - 9 = -5

Q3(i) 

Given the series 3, 1, -1,...

The first term is 3.

common difference = 1-3 =-2

(ii) -5, -1, 3, ..

First term = -5

Common difference = -1-(-5) = 4

(iii) 1/3, 5/3, 9/3 ...

First term = 1/3

Common difference = 5/3 - 1/3 = 4/3

(iv) 0.6, 1.7, 2.8,...

First term = 0.6

Common difference = 1.7-0.6 = 1.1

Q4. (i) 2, 4, 8, 16...

T1 = 2, T2 = 4, T3 = 8

T2-T1 = 4-2 = 2

T3-T2 = 8-4 = 4 

Since the differences are not the same, it is not A. P.

(ii) 2, 5/2, 3, 7/2

T1 = 2

T2 = 5/2, T3 = 3, T4 = 7/2

T2-T1 = 5/2-2 = 1/2

T3-T1 = 3-5/2 = 1/2 

Hence, it is in A.P. 

Next terms can be obtained by adding 1/2 as

2, 5/2, 3, 7/2, 4, 9/2, 5, 11/2, 

(iii) -1.2, -3.2, -5.2, -7.2 ....

T2-T1 = -3.2-(-1.2)  = -2

T3-T2 = -5.2-(-3.2) =-2

Since the difference between the terms is constant, hence it is an A.P. 

A few more terms: -1.2, -3.2, -5.2, -7.2, -9.2, -11.2,...

(iv) -10, -6, -2, 2 ...

T2-T1 = -6-(-10) = 4

T3-T2 = -5.2 -(-3.2) = 2

Hence, this is an A.P.

A few more terms 

-10, -6, -2, 2, 6, 10, 14, 18, ...

(v) 

(vi) 0.2, 0.22, 0.222, 0.222,0.2222, .....

T2-T1 = 0.22 - 0.20 = 0.02

T3 - T2 = 0.222 - 0.220 = 0.002

Here, the difference between the terms is not constant. So, it is not an A.P. 

(vii) 0, -4, -8, -12...

Difference between the terms—

-4-0 = -4

-8-(-4) = -4

T2-t1 is equal to T3-T2, so the first term is -4, T2 = -8, and d = -4.

The sequence can be written as 0, -4, -8, -12, -16, -20, -24....

(viii) -1/2, -1/2, -1/2...

On observing the series, it is obvious that d = 0. 

and a = -1/2 

Hence few more terms 

-1/2, -1/2, -1/2, -1/2, -1/2, .... 

(ix) 1, 3, 9, 27, ...

T2-T1 = 3-1 = 2

T3-T2 = 9-3 = 6. Since the difference between the terms is not constant, it is not an A.P. 

(x) a, 2a, 3a, 4a, .........

T2-T1 = 2a-a = a

T3-T2 = 3a-2a = a

The difference is constant and equals a

So, a few more terms are a, 2a, 3a, 4a, 5a, 6a...

(xi) a, a^2, a^3, a^4...

T2-T1 = a^2-a = a(a-1)

T3- T2 = a^3 - a^2 = a^2(a-1)

Hence, the difference is not constant. It is not an A.P.

(xii) 

(xiii)

(xiv) 1^2 = 1, 3^2 = 9, 5^2 = 25, 7^2 = 49...

T2 - T1 = 9 - 1 = 8

T3-T2 = 25-9 = 16

The difference is not the same; this is not an A.P. 

(xv)

 1^2 = 1

5^2 = 25

7^2 = 49 

73

T2 -T1 = 25-1 =24

T3- T2 = 49-25 = 24

The difference between the terms is the same; hence, it is an A.P. 

few more terms

1, 25, 49, 73, 97, 121, 145, ....