Chapter 5: Arithmetic Progressions

(A₁) AP is 2, – 2, – 6, – 10, ….

So common difference is simply

a₂ – a₁ = – 2 – 2 = – 4 = (B₃)

(A₂) Given

First term, a = – 18

No of terms, n = 10

Last term, a_n = 0

By using the nth term formula

a_{n =} a + (n – 1)d

0 = – 18 + (10 – 1)d

18 = 9d

d = 2 = (B₅)

(A₃) Given

First term, a = 0

Tenth term, a₁₀ = 6

By using the nth term formula

a_{n =} a + (n – 1)d

a₁₀ = a + 9d

6 = 0 + 9d

d = 2/3 = (B₆)

(A₄) Let the first term be a and common difference be d

Given that

a₂ = 13

a₄ = 3

a₂ – a₄ = 10

a + d – (a + 3d) = 10

d – 3d = 10

– 2d = 10

d = – 5= (B₁)

Here,

a₁ = 0

a₂ = ¼

a₃ = ½

a₄ = ¾

a₂ – a₁ = ¼ – 0 = ¼

a₃ – a₂ = ½ – ¼ = ¼

a₄ – a₃ = ¾ – ½ = ¼

Since, difference of successive terms are equal,

Hence, 0, 1/4, 1/2, 3/4… is an AP with common difference ¼.

Therefore, the next three term will be,

¾ + ¼ , ¾ + 2(¼), ¾ + 3(¼)

1, 5/4 , 3/2

(ii) 5, 14/3, 13/3, 4…

Here,

a₁ = 5

a₂ = 14/3

a₃ = 13/3

a₄ = 4

a₂ – a₁ = 14/3 – 5 = -1/3

a₃ – a₂ = 13/3 – 14/3 = -1/3

a₄ – a₃ = 4 – 13/3 = -1/3

Since, difference of successive terms are equal,

Hence, 5, 14/3, 13/3, 4… is an AP with common difference -1/3.

Therefore, the next three term will be,

4 + (-1/3), 4 + 2(-1/3), 4 + 3(-1/3)

11/3 , 10/3, 3

(iii) √3 , 2√3, 3√3,…

Here,

a₁ = √3

a₂ = 2√3

a₃ = 3√3

a₄ = 4√3

a₂ – a₁ = 2√3 – √3 = √3

a₃ – a₂ = 3√3 – 2√3= √3

a₄ – a₃ = 4√3 – 3√3= √3

Since, difference of successive terms are equal,

Hence, √3 , 2√3, 3√3,… is an AP with common difference √3.

Therefore, the next three term will be,

4√3 + √3, 4√3 + 2√3, 4√3 + 3√3

5√3, 6√3, 7√3

(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), …

Here

a₁ = a + b

a₂ = (a + 1) + b

a₃ = (a + 1) + (b + 1)

a₂ – a₁ = (a + 1) + b – (a + b) = 1

a₃ – a₂ = (a + 1) + (b + 1) – (a + 1) – b = 1

Since, difference of successive terms are equal,

Hence, a + b, (a + 1) + b, (a + 1) + (b + 1), … is an AP with common difference 1.

Therefore, the next three term will be,

(a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1(2), (a + 1) + (b + 1) + 1(3)

(a + 2) + (b + 1), (a + 2) + (b + 2), (a + 3) + (b + 2)

(v) a, 2a + 1, 3a + 2, 4a + 3,…

Here a₁ = a

a₂ = 2a + 1

a₃ = 3a + 2

a₄ = 4a + 3

a₂ – a₁ = (2a + 1) – (a) = a + 1

a₃ – a₂ = (3a + 2) – (2a + 1) = a + 1

a₄ – a₃ = (4a + 3) – (3a+2) = a + 1

Since, difference of successive terms are equal,

Hence, a, 2a + 1, 3a + 2, 4a + 3,… is an AP with common difference a+1.

Therefore, the next three term will be,

4a + 3 +(a + 1), 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1)

5a + 4, 6a + 5, 7a + 6

(i) a =1/2, d = -1/6

We know that,

First three terms of AP are :

a, a + d, a + 2d

½, ½ + (-1/6), ½ + 2 (-1/6)

½, 1/3, 1/6

(ii) a = –5, d = –3

We know that,

First three terms of AP are :

a, a + d, a + 2d

-5, – 5 + 1 (- 3), – 5 + 2 (- 3)

– 5, – 8, – 11

(iii) a = √2 , d = 1/√2

We know that,

First three terms of AP are :

a, a + d, a + 2d

√2, √2+1/√2, √2+2/√2

√2, 3/√2, 4/√2

For a, 7, b, 23, c… to be in AP

it has to satisfy the condition,

a₅ – a₄ = a₄ – a₃ = a₃ – a₂ = a₂ – a₁ = d

Where d is thecommon difference

7 – a = b – 7 = 23 – b = c – 23 …(1)

Let us equate,

b – 7 = 23 – b

2b = 30

b = 15 (eqn 1)

And,

7 – a = b – 7

From eqn 1

7 – a = 15 – 7

a = – 1

And,

c – 23 = 23 – b

c – 23 = 23 – 15

c – 23 = 8

c = 31

So a = – 1

b = 15

c = 31

Then, we can say that, the sequence – 1, 7, 15, 23, 31 is an AP

We know that,

The first term of an AP = a

And, the common difference = d.

According to the question,

5^th term, a₅ = 19

Using the n^th term formula,

a_n = a + (n – 1)d

We get,

a + 4d = 19

a = 19 – 4d …(1)

Also,

13^th term – 8^th term = 20

a + 12d – (a + 7d) = 20

5d = 20

d = 4

Substituting d = 4 in equation 1,

We get,

a = 19 – 4(4)

a = 3

Then, the AP becomes,

3, 3 + 4 , 3 + 2(4),…

3, 7, 11,…