Exercise 5.1

(i)

The taxi fare after each km when the fare is ₹15 for the first km and ₹ 8 for each additional km.

Solution

Taxi fare for 1 km = ₹ (a1)

Taxi fare for 2 km = 15 + = ₹ (a2)

Taxi fare for 3 km = 15 + 8 + 8 = ₹ (a3)

(a2) - (a1) = ₹ (23 - 15) = ₹

(a3) - (a2) = ₹ (31 - 23) = ₹

We see that the difference is constant between the terms.

So, this forms an AP with the first term as 15 and the common difference 8.

(ii)

The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Solution

Let the amount of air in the cylinder be x.

So a1 =

After first time removal, a2 = x - (x4) = 3x4

After second time removal,

a3 = (3x4) - (14) (3x4)

= (3x4) - (3x16)

= (12x - 3x)16

= 9x16916

After third time removal

a4 = (9x16) - (14) (9x16)

= (9x16) - (9x64)

= (36x - 9x)64

= 27x6464x27

Now,

a2 - a1 = (3x4) - x

= (3x - 4x)4

= −x4x4

a3 - a2 = (9x16) - (3x4)

= (9x - 12x)16

= -3x163x16

(a3 - a2) ≠= (a2 - a1)

Since the difference between the terms is not the same, this is not forming an AP.

(iii)

The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.

Solution

Cost of digging the well after 1 meter = ₹ (a1)

Cost of digging the well after 2 meters = ₹ 150 + = ₹ (a2)

Cost of digging the well after 3 meters = ₹ 150 + 50 + 50 = ₹ (a3)

(a2 - a1) = 200 - 150 =

(a3 - a2) = 250 - 200 =

(a2 - a1) =≠ (a3 - a2)

We see that the difference is constant between the terms.

So, this froms an AP with the first term as ₹ 150 and the common difference is ₹ 50.

(iv)

The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.

Solution

Amount present when the sum is P at an interest rate of r % after n years is

A = P(1 + r100)n

P =

r = %

For first year, n = , (a1) = 10000 (1 + 8100)

For second year, n = , (a2) = 10000 1+81002

For third year, n = , (a3) = 10000 1+81003

For fourth year, n = , (a₄) = 10000 1+81004

a2 - a1 = [10000 1+81002] - [10000 (1 + 8100)]

= [10000 (1 + 8100)] [(1 + 8100) - )]

= [10000 (1 + 8100)] (8100)

a3 - a2

= [10000 1+81003] - [10000 1+81002]

= [10000 1+81002] [1 + 8100 - ]

= [10000 1+81002 ] (8100)

(a3 - a2)≠= (a2 - a1)

Since the difference between the terms is not the same thus, the amount will not form an AP.

(i)

a = 10, d = 10.

Solution

a = , d =

First term, a = 10

Second term, a + d = 10 + 10 =

Third term, a + 2d = 10 + =

Fourth term, a + 3d = 10 + =

The first four terms of the AP are , , , and .

(ii)

a = - 2, d = 0

Solution

First term, a =

Second term, a + d = - 2 + 0 =

Third term, a + 2d = - 2 + 0 =

Fourth term, a + 3d = - 2 + 0 =

The first four terms of the AP are - 2, - 2, - 2, and -2.

(iii)

a = 4, d = - 3

Solution

First term, a =

Second term, a + d = 4 + (- 3) =

Third term, a + 2d = 4 + 2(- 3) = 4 - 6 =

Fourth term, a + 3d = 4 + 3(- 3) = 4 - 9 =

The first four terms of the AP are 4, 1, - 2, - 5.

(iv)

a = - 1, d = 12

Solution

First term, a =

Second term, a + d = - 1 + 12 = −1212

Third term, a + 2d = - 1 + 2(12) = - 1 + 1 =

Fourth term, a + 3d = - 1 + 3(12) = - 1 + 32 = 12−12

The first four terms of the AP are - 1, −12, 0, 12

(v)

a = - 1.25, d = - 0.25

Solution

First term, a =

Second term, a + d = -1.25 + (-0.25) = - 1.25 - 0.25 =

Third term, a + 2d = - 1.25 + 2 (- 0.25) = - 1.25 - 0.50 =

Fourth term, a + 3d = - 1.25 + 3 (-0.25) = - 1.25 - 0.75 =

The first four terms of the AP are - 1.25, - 1.5, - 1.75, and - 2.

(i)

3, 1, – 1, – 3, . . .

Solution

3, 1, - 1, - 3,… are in AP

First term, a1 =

Common difference, d = a2 - a1

d = 1 - 3 =

Thus, the first term is and the common difference is .

(ii)

– 5, – 1, 3, 7, . . .

Solution

-5, -1, 3, 7,… are in AP

First term, a1 =

Common difference, d = a2 - a1

d = - 1 - (- 5) = - 1 + 5 =

Thus, the first term is and the common difference is .

(iii)

13, 53, 93, 133,…

Solution

13, 53, 93, 133,… are in AP

First term, a1 = 13−13

Common difference, d = a2 - a1

d = 53 - 13 = 4334

Thus, the first term is 13 and the common difference is 43.

(iv)

0.6, 1.7, 2.8, 3.9, . . .

Solution

0.6, 1.7, 2.8, 3.9,… are in AP

First term, a1 =

Common difference, d = a2 - a1

d = 1.7 - 0.6 =

Thus, the first term is and the common difference is .

(i)

2, 4, 8, 16, ...

Solution

2, 4, 8, 16......

First term, a1 =

Common difference d = a2 - a1 = 4 - 2 =

Common difference d = a3 - a2 = 8 - 4 =

(a3 - a2) ≠= (a2 - a1)

So, 2, 4, 8, 16, ... are not in AP, because the common difference is not equal.

(ii)

2, 52 ,3, 72, ...

Solution

2, 52 ,3, 72, ...

First term, a1 =

Common difference d = a2 - a1 = 52 - 2 = (5 - 4)2 = 12−12

Common difference d = a3 - a2= 3 - 52 = (6 - 5)2 =12−12

Since a3 - a2 = a2 - a1.

2, 52 ,3, 72 forms an AP and common difference is 12

The next three terms are:

Fifth term = a1 + 4d = 2 + 4 × 12 = 2 + 2 =

Sixth term = a1 + 5d = 2 + 5 × 12 = 2 + 52 = 4 + 52 = 9229

Seventh term = a1+ 6d = 2 + 6 × 12 =

2, 52 ,3, 72, ... forms an AP and the common difference is 12. The next three terms are 4, 92, 5

(iii)

- 1.2, - 3.2, - 5.2, - 7.2, ...

Solution

-1.2, - 3.2, - 5.2, - 7.2, ...

First term, a1=

Common difference d = a2 - a1= -3.2 - (-1.2) = -3.2 + 1.2 =

Common difference d = a3 - a2= - 5.2 - (-3.2) = - 5.2 + 3.2 =

Since a3 - a2 = a2 - a1, it forms an AP.

Fifth term = a1 + 4d = - 1.2 + 4(- 2) = -1.2 - 8 =

Sixth term = a1 + 5d = - 1.2 + 5(- 2) = - 1.2 - =

Seventh term = a1 + 6d = - 1.2 + 6(- 2) = - 1.2 - =

- 1.2, - 3.2, - 5.2, - 7.2, ... forms an AP with common difference - 2. The next three terms of AP are - 9.2, - 11.2, - 13.2

(iv)

- 10, - 6, - 2, 2, ...

Solution

-10, - 6, - 2, 2, ...

First term, a1 =

Common difference d = a2 - a1 = - 6 - (- 10) = - 6 + 10 =

Common difference d = a3 - a2 = - 2 - (- 6) = - 2 + 6 =

Since a3 - a2 = a2 - a1, it forms an AP.

Fifth term = a1 + 4d = - 10 + 16 =

Sixth term = a1 + 5d = - 10 + 20 =

Seventh term = a1 + 6d = - 10 + 24 =

- 10, -6, - 2, 2 forms an AP with common difference 4 and next three terms are 6, 10, 14.

(v)

3, 3 + √2, 3 + 2√2, 3 + 3√2, ...

Solution

3, 3 + √2, 3 + 2√2, 3 + 3√2, ...

First term, a1 =

Common difference d = a2 - a1 = 3 + √2 - 3 = √22

Common difference d = a3 - a2 = 3 + 2√2 - (3 + √2) = 3 + 2√2 - 3 - √2

= √22

Since a3 - a2 = a2 - a1, 3, 3 + √2, 3 + 2√2, 3 + 3√2, ... forms an AP.

So, 3, 3 + √2, 3 + 2√2, 3 + 3√2 forms an AP with common difference .

Next three terms are

Fifth term = a1 + 4d = 3 + 4 × √2 = 3 + √2

Sixth term = a1 + 5d = 3 + 5 × √2 = 3 + √2

Seventh term = a1 + 6d = 3 + 6 × √2 = 3 + √2

3, 3 + √2, 3 + 2√2, 3 + 3√2 forms an AP with common difference √22 and next three terms are 3 + 4√2, 3 + 5√2, 3 + 6√2

(vi)

0.2, 0.22, 0.222, 0.2222, ...

Solution

0.2, 0.22, 0.222, 0.2222, ...

First term, a1 =

Common difference d = a2 - a1 = 0.22 - 0.2 =

Common difference d = a3 - a2= 0.222 - 0.220 =

Since (a₃ - a₂) ≠= (a₂ - a₁), 0.2, 0.22, 0.222, 0.2222, ... do not form an AP.

So, the given list of numbers does not form an AP.

(vii)

0, - 4, - 8, - 12, ...

Solution

0, - 4, - 8, - 12, ...

First term, a1 =

Common difference d = a2 - a1 = - 4 - 0 =

Common difference d = a3 - a2 = - 8 - (- 4) = - 8 + 4 =

Since a3 - a2 = a2 - a1, it forms an AP.

Fifth term = a1 + 4d = 0 + 4(- 4) =

Sixth term = a1 + 5d = 0 + 5(- 4) =

Seventh term = a1 + 6d = 0 + 6(- 4) =

0, - 4, - 8, - 12 forms an AP with a common difference of - 4. The next three terms are -16, -20, -24.

(viii)

−12, −12, −12, −12,....

Solution

−12, −12, −12, −12,....

First term, a1 =

Common difference d = a2 - a1 = −12 - (−12)= −12 + 12 =

Common difference d = a3 - a2= −12 - (−12)= −12 + 12 =

Since a3 - a2 = a2 - a1, it forms an AP.

Fifth term = a1 + 4d = −12 + 4 (0) = −1212

Sixth term = a1 + 5d = −12 + 5 (0) = −1212

Seventh term = a1 + 6d = −12 + 6 (0) = −1212

−12, −12, −12, −12 forms an AP with a common difference d = 0. The next three terms are −12, −12, −12.

(ix)

a, 2a, 3a, 4a, ...

Solution

a, 2a, 3a, 4a,.....

First term, a1 =

Common difference d = a2 - a1= 2a - a =

Common difference d = a3 - a2 = 3a - 2a =

Since a3 - a2 = a2 - a1, it forms an AP.

Fifth term = a1 + 4d = a + 4a =

Sixth term = a1 + 5d = a + 5a =

Seventh term = a1 + 6d = a + 6a =

a, 2a, 3a, 4a forms an AP with a common difference d = a. The next three terms are 5a, 6a, 7a.

(x)

12, 32, 52, 72, ....

Solution

First term, a1= 12−12

Common difference d = a2 - a1 = 9 - 1 =

Common difference d = a3 - a2 = 25 - 9 =

Since a2 - a1 ≠= a3 - a2, the given list of numbers does not form an AP.

(xi)

a, a2, a3, a4, ...

Solution

a, a2, a3, a4, ...

First term, a1 =

Common difference d = a2 - a1 = a2 - a = a (a - )

Common difference d = a3 - a2 = a3 - a2 = a2 (a - )

Sincea2 - a1 ≠= a3 - a2, the given list of numbers does not form an AP.