Arithmetic Progression — Multiple Choice Questions

The 7th term of the given Arithmetic Progression (A.P.):

$\dfrac{1}{a}, \Big(\dfrac{1}{a} + 1\Big), \Big(\dfrac{1}{a} + 2\Big), ......$ is :

1. $\Big(\dfrac{1}{a} + 6\Big)$

2. $\Big(\dfrac{1}{a} + 7\Big)$

3. $\Big(\dfrac{1}{a} + 8\Big)$

4. $\Big(\dfrac{1}{a} + 7^7\Big)$

Answer

In the A.P. :

$\dfrac{1}{a}, \Big(\dfrac{1}{a} + 1\Big), \Big(\dfrac{1}{a} + 2\Big), ......$

First term (x) = $\dfrac{1}{a}$

Common difference (d) = $\dfrac{1}{a} + 1 - \dfrac{1}{a}$ = 1.

x₇ = x + (7 - 1)d

= $\dfrac{1}{a} + 6 \times 1$

= $\Big(\dfrac{1}{a} + 6\Big)$.

Hence, Option 1 is the correct option.

The nth term from the end of an A.P., whose first term, last term and common difference are a, l and d respectively, is given by:

1. l + (n + 1)d

2. l − (n − 1)d

3. l + (n − 1)d

4. l − (n + 1)d

Answer

Since, we need to calculate term from end.

So, first term will be equal to last term and common difference will be negative of the original difference.

∴ T_end = l + (n - 1)(-d)

= l - (n - 1)d.

Hence, option 2 is the correct option.

The sum of first n natural numbers is:

1. $\dfrac{n(n − 1)}{2}$

2. $\dfrac{n(n + 1)}{2}$

3. $\dfrac{n(n + 2)}{2}$

4. $\dfrac{n(n − 2)}{2}$

Answer

The sequence of natural numbers 1, 2, 3,....., n.

a = 1

l = n

no. of terms = n

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

= $\dfrac{n}{2}$ (1 + n)

= $\dfrac{n(n + 1)}{2}$

Hence, option 2 is the correct option.

The sum of first 50 natural numbers is:

1. 1050

2. 1175

3. 1225

4. 1275

Answer

Sequence : 1, 2, 3, ......, 50.

First term (a) = 1

Common difference (d) = 2 - 1 = 1

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

$= \dfrac{50}{2} \times [2 \times 1 + (50 - 1) \times 1] \\[1em] = 25 \times [2 + 49] \\[1em] = 25 \times 51 \\[1em] = 1275.$

Hence, option 4 is the correct option.

The nth term of an Arithmetic Progression (A.P.) is 2n + 5. The 10th term is :

1. 7

2. 15

3. 25

4. 45

Answer

Given,

nth term of A.P. :

∴ T_n = 2n + 5

⇒ T₁₀ = 2(10) + 5

= 20 + 5

= 25.

Hence, option 3 is the correct option.

The first term of an A.P. is 7 and the common difference is 3. The general term of the A.P. is:

1. T_n = 3n − 4

2. T_n = 2n + 5

3. T_n = 3n + 4

4. None of these

Answer

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

Given,

a = 7, d = 3

T_n = 7 + (n - 1)3

= 7 + 3n - 3

= 3n + 4.

Hence, option 3 is the correct option.

The 24th term of the A.P. −1, 3, 7, 11, … is:

1. 83

2. 87

3. 91

4. 95

Answer

The given Arithmetic Progression (A.P.) is

-1, 3, 7, 11,.....

a = -1

d = 3 - (-1) = 4

n = 24

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₂₄ = -1 + (24 - 1)(4)

= -1 + (23)4

= -1 + 92

= 91.

Hence, option 3 is the correct option.

If 70, 75, 80, 85 are the first four terms of an arithmetic progression, then the 10th term is:

1. 35

2. 25

3. 115

4. 105

Answer

The given Arithmetic Progression (A.P.) is

70, 75, 80, 85, ....

a = 70

d = 75 - 70 = 5

n = 10

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₁₀ = 70 + (10 - 1)5

= 70 + (9)5

= 70 + 45

= 115.

Hence, option 3 is the correct option.

The A.P. 6, 13, 20, …, 216 has 31 terms. The middle term of the A.P. is :

1. 91

2. 97

3. 107

4. 111

Answer

Given,

A.P. 6, 13, 20, …, 216

n = 31

Middle term of A.P. = $\dfrac{n + 1}{2}$

= $\dfrac{31 + 1}{2}$

= $\dfrac{32}{2}$

= 16th term.

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

Now we have,

a = 6

d = 13 - 6 = 7

n = 16

⇒ T₁₆ = 6 + (16 - 1)(7)

= 6 + (15)7

= 6 + 105

= 111.

Hence, option 4 is the correct option.

Which term of the A.P. 7, 13, 19, 25,..... is 241?

1. 40th

2. 36th

3. 44th

4. 45th

Answer

Given,

A.P. : 7, 13, 19, 25,.......

a = 7

d = 13 - 7 = 6

Let nth term be 241.

T_n = 241

⇒ a + (n - 1)d = 241

⇒ 241 = 7 + (n - 1)6

⇒ 241 - 7 = (n - 1)6

⇒ 234 = (n - 1)6

⇒ $\dfrac{234}{6}$ = n - 1

⇒ n - 1 = 39

⇒ n = 39 + 1

⇒ n = 40.

Hence, option 1 is the correct option.

Which term of the A.P. 11, 8, 5, 2, ..... is −148 ?

1. 52nd

2. 54th

3. 55th

4. 57th

Answer

Given,

A.P. : 11, 8, 5, 2, .......

a = 11

d = 8 - 11 = -3

Let nth term of the A.P. be -148.

T_n = -148

⇒ a + (n - 1)d = -148

⇒ -148 = 11 + (n - 1)(-3)

⇒ -148 - 11 = (n - 1)(-3)

⇒ -159 = (n - 1)(-3)

⇒ $\dfrac{-159}{-3}$ = n - 1

⇒ n - 1 = 53

⇒ n = 53 + 1

⇒ n = 54.

Hence, option 2 is the correct option.

How many three-digit numbers are divisible by 7?

1. 128

2. 124

3. 136

4. 132

Answer

The three-digit numbers divisible by 7 form an Arithmetic Progression (A.P.) :

105, 112, 119, ......,994.

a = 105

l = 994

d = 7

Let no. of terms be n.

∴ T_n = a + (n - 1)d

⇒ 994 = 105 + (n - 1)7

⇒ 994 - 105 = (n - 1)7

⇒ 889 = (n - 1)7

⇒ $\dfrac{889}{7}$ = n - 1

⇒ 127 = n - 1

⇒ n = 127 + 1

⇒ n = 128.

Hence, option 1 is the correct option.

How many numbers lying between 20 and 200 are divisible by 4?

1. 43

2. 44

3. 45

4. 46

Answer

The numbers divisible by 4 lying between 20 and 200 form an Arithmetic Progression (A.P.).

24, 28, ....., 196.

a = 24

l = 196

d = 4

Let no. of terms be n.

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ 196 = 24 + (n - 1)4

⇒ 196 - 24 = (n - 1)4

⇒ 172 = (n - 1)4

⇒ $\dfrac{172}{4}$ = n - 1

⇒ 43 = n - 1

⇒ n = 43 + 1

⇒ n = 44.

Hence, option 2 is the correct option.

The common difference of the A.P. $\dfrac{1}{k}, \dfrac{1 - k}{k}, \dfrac{1 - 2k}{k}$, ...... is :

1. −k

2. k

3. 1

4. −1

Answer

Given,

A.P. : $\dfrac{1}{k}, \dfrac{1 - k}{k}, \dfrac{1 - 2k}{k},$ .........

a = $\dfrac{1}{k}$

$\Rightarrow d = \dfrac{1 - k}{k} - \dfrac{1}{k} \\[1em] \Rightarrow d = \dfrac{1 - k - 1}{k} \\[1em] \Rightarrow d = \dfrac{-k}{k} \\[1em] \Rightarrow d = -1.$

Hence, option 4 is the correct option.

If the nth term of an A.P. is given by (3n + 2), then the sum of its first three terms is :

1. 21

2. 24

3. 27

4. 32

Answer

The nth term of the A.P. is given by,

T_n = 3n + 2

T₁ = 3(1) + 2 = 5

T₂ = 3(2) + 2 = 8

T₃ = 3(3) + 2 = 11

Sum = 5 + 8 + 11 = 24.

Hence, option 2 is the correct option.

The last term of the A.P., 5, 12, 19, ..... having 60 terms is :

1. 406

2. 412

3. 416

4. 418

Answer

Given,

A.P. : 5, 12, 19, .......

a = 5

d = 12 - 5 = 7

n = 60

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₆₀ = 5 + (60 - 1)7

= 5 + (59) × 7

= 5 + 413

= 418.

Hence, option 4 is the correct option.

The common difference of the A.P. : $\dfrac{1}{3}, \dfrac{(1 − 3p)}{3}, \dfrac{(1 − 6p)}{3}$, ... is:

1. p

2. −p

3. $\dfrac{1}{3}$

4. $\dfrac{-1}{3}$

Answer

In the above A.P.,

$\Rightarrow d = \dfrac{(1 − 3p)}{3} - \dfrac{1}{3} \\[1em] \Rightarrow d = \dfrac{(1 − 3p) - 1}{3} \\[1em] \Rightarrow d = \dfrac{-3p}{3} \\[1em] \Rightarrow d = -p.$

Hence, option 2 is the correct option.

The next term of the A.P. $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}$, ... is:

1. $\sqrt{72}$

2. $\sqrt{68}$

3. $\sqrt{75}$

4. $\sqrt{56}$

Answer

In the above A.P. next terms is the 5th term.

a = $\sqrt{3}$

n = 5

$\Rightarrow d = \sqrt{12} - \sqrt{3} \\[1em] \Rightarrow d = \sqrt{4 \times 3} - \sqrt3 \\[1em] \Rightarrow d = 2\sqrt{3} - \sqrt3 \\[1em] \Rightarrow d = \sqrt3.$

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

$\Rightarrow T_5 = \sqrt3 + (5 - 1)\sqrt3 \\[1em] = \sqrt3 + 4\sqrt3 \\[1em] = 5\sqrt3 \\[1em] = \sqrt{5^2 \times 3} \\[1em] = \sqrt{75}.$

Hence, option 3 is the correct option.

The first term of an A.P. is p and its common difference is q. The 10th term of the A.P. is :

1. p − 9q

2. p + 10q

3. p − 10q

4. p + 9q

Answer

Given,

First term = p

Common difference = q

n = 10

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₁₀ = p + (10 - 1)q

= p + 9q.

Hence, option 4 is the correct option.

The nth term of the A.P. $\dfrac {1}{m}, \dfrac{(1 + m)}{m}, \dfrac{(1 + 2m)}{m}$, ... is:

1. $\dfrac{m(n − 1) + 1}{m}$

2. $\dfrac{m(n + 1) − 1}{m}$

3. $\dfrac{m(n − 1) − 1}{m}$

4. None of these

Answer

Given,

a = $\dfrac{1}{m}$

$\Rightarrow d = \dfrac{(1 + m)}{m} - \dfrac {1}{m} \\[1em] \Rightarrow d = \dfrac{(1 + m) - 1}{m} \\[1em] \Rightarrow d = \dfrac{m}{m} \\[1em] \Rightarrow d = 1.$

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

$\Rightarrow T_n = \dfrac{1}{m} + (n - 1) \times 1 \\[1em] = \dfrac{1 + m(n - 1)}{m} \\[1em] = \dfrac{m(n - 1) + 1}{m}.$

Hence, option 1 is the correct option.

For what value of p are 2p + 1, 13, 5p − 3 three consecutive terms of an A.P. ?

1. 0

2. 1

3. 2

4. 4

Answer

Given,

2p + 1, 13, 5p − 3

In an A.P., the middle term is the average of the first and third terms.

$\Rightarrow 13 = \dfrac{(2p + 1) + (5p - 3)}{2} \\[1em] \Rightarrow 13 = \dfrac{7p - 2}{2} \\[1em] \Rightarrow 13 \times 2 = 7p - 2 \\[1em] \Rightarrow 26 = 7p - 2 \\[1em] \Rightarrow 26 + 2 = 7p \\[1em] \Rightarrow 28 = 7p \\[1em] \Rightarrow 7p = 28 \\[1em] \Rightarrow p = \dfrac{28}{7} \\[1em] \Rightarrow p = 4.$

Hence, option 4 is the correct option.

For what value of k will 2k + 1, 3k + 3, and 5k − 1 be three consecutive terms of an A.P.?

1. 1

2. 2

3. 4

4. 6

Answer

We are given three consecutive terms of an A.P.:

2k + 1, 3k + 3, 5k − 1

In an arithmetic progression, the difference between consecutive terms is the same.

⇒ (3k + 3) - (2k + 1) = (5k - 1) - (3k + 3)

⇒ 3k + 3 - 2k - 1 = 5k - 1 - 3k - 3

⇒ k + 2 = 2k - 4

⇒ 2 + 4 = 2k - k

⇒ k = 6.

Hence, option 4 is the correct option.

If the sum of first n terms of an A.P. is S_n = 5n² + 3n, then its common difference is:

1. 8

2. 10

3. 18

4. 26

Answer

Given,

S_n = 5n² + 3n

We know that,

nth term of A.P. :

∴ T_n = S_n - S_{n - 1}

= 5n² + 3n - [5(n - 1)² + 3(n - 1)]

= 5n² + 3n - [5(n² - 2n + 1) + 3n - 3]

= 5n² + 3n - [5n² - 10n + 5 + 3n - 3]

= 5n² + 3n - 5n² + 10n - 5 - 3n + 3

= 10n - 2

d = a_n - a_{n - 1}

= 10n - 2 - [10(n - 1) - 2]

= 10n - 2 - [10n - 10 - 2]

= 10n - 2 - 10n + 10 + 2

= 10.

Hence, option 2 is the correct option.

The first and the last terms of an A.P. are 1 and 11 respectively. If the sum of its terms is 36, then the number of terms is :

1. 6

2. 7

3. 8

4. 9

Answer

Given,

a = 1

l = 11

Let no. of terms be n.

S_n = 36

We know that,

⇒ S_n = $\dfrac{n}{2}$ (a + l)

⇒ 36 = $\dfrac{n}{2}$ (1 + 11)

⇒ 36 × 2 = n(12)

⇒ n = $\dfrac{36 \times 2}{12}$

⇒ n = 6.

Hence, option 1 is the correct option.

The sum of first 40 positive integers divisible by 6 is :

1. 2460

2. 3640

3. 4920

4. 4860

Answer

Sequence :

6, 12, 18, ......., upto 40th term.

The above sequence is an A.P. with,

a = 6

d = 6

n = 40

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₄₀ = $\dfrac{40}{2}$ [2(6) + (40 - 1)6]

= 20[12 + (39)6]

= 20(12 + 234)

= 20 × (246)

= 4920.

Hence, option 3 is the correct option.

The sum of first 30 odd natural numbers is:

1. 800

2. 900

3. 729

4. 1249

Answer

1, 3, 5, 7, ......, 30th term.

The odd natural numbers form an Arithmetic Progression (A.P.) :

a = 1

d = 3 - 1 = 2

n = 30

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₃₀ = $\dfrac{30}{2}$ [2(1) + (30 - 1)2]

= 15[2 + (29)2]

= 15[2 + 58]

= 15 × (60)

= 900.

Hence, option 2 is the correct option.

The 7th term of an A.P. is 4 and its common difference is −4. The first term of the A.P. is :

1. 32

2. 28

3. 24

4. 36

Answer

Given,

a₇ = 4

d = -4

n = 7

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₇ = a + (7 - 1)(-4)

⇒ 4 = a + (6)(-4)

⇒ 4 = a - 24

⇒ 24 + 4 = a

⇒ a = 28.

Hence, option 2 is the correct option.

The sum of first n terms of an A.P. is (4n² + 2n). The nth term of the A.P. is :

1. (6n − 2)

2. (8n − 2)

3. (6n + 2)

4. (8n + 2)

Answer

We know that,

S_n = 4n² + 2n

T_n = S_n - S_{n - 1}

= 4n² + 2n - [4(n - 1)² + 2(n - 1)]

= 4n² + 2n - [4(n² - 2n + 1) + 2n - 2]

= 4n² + 2n - [4n² - 8n + 4 + 2n - 2]

= 4n² + 2n -[4n² - 6n + 2]

= 4n² + 2n - 4n² + 6n - 2

= 8n - 2.

Hence, option 2 is the correct option.

The first term of an A.P. is 7 and its 13th term is 35. The common difference of the A.P. is :

1. $\dfrac{5}{3}$

2. $\dfrac{7}{2}$

3. $\dfrac{8}{3}$

4. $\dfrac{7}{3}$

Answer

Given,

a = 7

⇒ a₁₃ = 35

⇒ a + (13 - 1)d = 35

⇒ 7 + 12d = 35

⇒ 12d = 35 - 7

⇒ 12d = 28

⇒ d = $\dfrac{28}{12}$

⇒ d = $\dfrac{7}{3}$.

Hence, option 4 is the correct option.

There are total 9 terms in an A.P. If the last term and the sum of all the terms are 28 and 144 respectively, then the first term is :

1. 1

2. 7

3. 4

4. 5

Answer

We know that,

S_n = $\dfrac{n}{2}$(a + l)

Given,

n = 9

l = 28

S_n = 144

⇒ S₉ = $\dfrac{9}{2}$(a + 28)

⇒ 144 = $\dfrac{9}{2}$(a + 28)

⇒ $\dfrac{144 \times 2}{9}$ = (a + 28)

⇒ 32 = (a + 28)

⇒ 32 - 28 = a

⇒ a = 4.

Hence, option 3 is the correct option.

The 5th term of an A.P. is −3 and its common difference is −4. The sum of its first 10 terms is :

1. −50

2. −40

3. −60

4. −30

Answer

Given,

a₅ = -3

d = -4

n = 10

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₅ = a + (5 - 1)(-4)

⇒ -3 = a + (4)(-4)

⇒ -3 = a - 16

⇒ a = 16 - 3

⇒ a = 13.

We know that,

⇒ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₀ = $\dfrac{10}{2}$ [2(13) + (10 - 1)(-4)]

= 5[26 + (9)(-4)]

= 5[26 - 36]

= 5(-10)

= -50.

Hence, option 1 is the correct option.

The 7th term of an A.P. is −1 and its 16th term is 17. The nth term of the A.P. is :

1. (3n + 12)

2. (2n − 5)

3. (3n + 5)

4. (2n − 15)

Answer

We know that,

a_n = a + (n - 1)d

The 7th term of an A.P. is −1.

a + 6d = -1 ......(1)

The 16th term of an A.P. is 17.

a + 15d = 17 ......(2)

Subtract Equation (1) from Equation (2) :

⇒ a + 15d - (a + 6d) = 17 - (-1)

⇒ a + 15d - a - 6d = 18

⇒ 9d = 18

⇒ d = $\dfrac{18}{9}$

⇒ d = 2.

Substituting d = 2 into Equation (1), we get :

⇒ a + 6(2) = -1

⇒ a + 12 = -1

⇒ a = -1 - 12

⇒ a = -13.

Now,

⇒ a_n = a + (n - 1)d

⇒ a_n = -13 + (n - 1)2

= -13 + 2n - 2

= 2n - 15.

Hence, option 4 is the correct option.

If the sum of first p terms of an A.P. is ap² + bp, then its common difference is :

1. a

2. 3a + b

3. a + b

4. 2a

Answer

Given,

⇒ S_p = ap² + bp

⇒ T_p = S_p - S_{p - 1}

= ap² + bp - [a(p - 1)² + b(p - 1)]

= ap² + bp - [a(p² - 2p + 1) + bp - b]

= ap² + bp - [ap² - 2ap + a + bp - b]

= ap² + bp - ap² + 2ap - a - bp + b

= 2ap - a + b.

d = T_p - T_{p - 1}

= 2ap - a + b - [2a(p - 1) - a + b]

= 2ap - a + b - [2ap - 2a - a + b]

= 2ap - a + b - 2ap + 2a + a - b

= 2a.

Hence, option 4 is the correct option.

The first term of an A.P. is a and nth term is b, then the common difference of the A.P. is :

1. $\dfrac{b − a}{n}$

2. $\dfrac{b − a}{n − 1}$

3. $\dfrac{b + a}{n − 1}$

4. $\dfrac{b − a}{n + 1}$

Answer

Given,

First term = a

nth term = b

We know that,

a_n = a + (n - 1)d

⇒ b = a + (n - 1)d

⇒ b - a = (n - 1)d

⇒ d = $\dfrac{b - a}{n - 1}$.

Hence, option 2 is the correct option.

The first three terms of an A.P. are 3y − 1, 3y + 5 and 5y + 1 respectively. Then, the value of y is :

1. 1

2. 5

3. 8

4. 3

Answer

Given,

The first three terms of an A.P. are 3y − 1, 3y + 5 and 5y + 1.

The difference between consecutive terms must be equal.

⇒ 3y + 5 - (3y - 1) = 5y + 1 - (3y + 5)

⇒ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5

⇒ 6 = 2y - 4

⇒ 2y = 6 + 4

⇒ 2y = 10

⇒ y = $\dfrac{10}{2}$

⇒ y = 5.

Hence, option 2 is the correct option.

The 8th term from the end of the A.P. 7, 10, 13, ..., 184 is:

1. 157

2. 160

3. 163

4. 166

Answer

Given,

a = 7

d = 10 - 7 = 3

a_n = 184

We know that,

⇒ a_n = a + (n - 1)d

⇒ a_n = 7 + (n - 1)(3)

⇒ 184 = 7 + (n - 1)(3)

⇒ 184 - 7 = (n - 1)(3)

⇒ 177 = (n - 1)(3)

⇒ $\dfrac{177}{3}$ = (n - 1)

⇒ n - 1 = 59

⇒ n = 59 + 1

⇒ n = 60.

The 8th term from the end is the (n - 8 + 1)th term from the beginning.

= 60 - 8 + 1

= 53.

⇒ a₅₃ = 7 + (53 - 1)3

= 7 + (52)(3)

= 7 + 156

= 163.

Hence, option 3 is the correct option.

If a = 3, n = 8 and S_n = 192, then the common difference of the A.P. is:

1. 4

2. 5

3. 6

4. 7

Answer

a = 3

n = 8

S_n = 192

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ 192 = $\dfrac{8}{2}$ [2(3) + (8 - 1)d]

⇒ 192 = 4[6 + 7d]

⇒ $\dfrac{192}{4}$ = [6 + 7d]

⇒ 48 = [6 + 7d]

⇒ 48 - 6 = 7d

⇒ 42 = 7d

⇒ d = $\dfrac{42}{7}$

⇒ d = 6.

Hence, option 3 is the correct option.

If the sum of n terms of an arithmetic progression S_n = n² - n, then the third term of the series is :

1. 2

2. 4

3. 6

4. 9

Answer

Given,

Sum of n terms of an arithmetic progression S_n = n² - n.

S₁ = 1² - 1 = 0,

S₂ = 2² - 2 = 4 - 2 = 2,

S₃ = 3² - 3 = 9 - 3 = 6.

Sum upto first term = First term = 0.

Given, sum upto 2 terms = 2 and first term = 0, second term = 2.

Sum upto third term = 6

∴ First term + Second term + Third term = 6

⇒ 0 + 2 + Third term = 6

⇒ Third term = 6 - 2 = 4.

Hence, Option 2 is the correct option.