Arithmetic Progression — Assertion-Reason Type Questions

Assertion (A): The sum of first n terms of the A.P. −1, 5, 11, ... is 3n² − 4n.

Reason (R): The sum of first n terms of an A.P. is given by S_n = $\dfrac{n}{2}$ [2a + (n − 1)d].

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

A.P. : -1, 5, 11, ......

Given,

a = -1

d = 5 - (-1) = 6

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n − 1)d]

⇒ S_n = $\dfrac{n}{2}$ [2(-1) + (n − 1)6]

= $\dfrac{n}{2}$ [-2 + (6n − 6)]

= $\dfrac{n}{2}$ (6n − 8)

= $\dfrac{n}{2} \times$ 2(3n − 4)

= n(3n - 4)

= 3n² - 4n.

∴ Assertion (A) is true.

The standard and correct formula for the sum of the first n terms of an A.P.

S_n = $\dfrac{n}{2}$[2a + (n − 1)d]

∴ Reason (R) is true.

Both A and R are true, and R is the correct explanation of A.

Hence, option 1 is the correct option.

Assertion (A): The 10th term from the end of the A.P. 17, 14, 11, ... −40 is −11.

Reason (R): The nth term of an A.P. is given by t_n = a + (n − 1)d.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Given,

a = 17

a_n = -40

d = 14 - 17 = -3

We know that,

⇒ a_n = a + (n - 1)d

⇒ -40 = 17 + (n - 1)(-3)

⇒ -40 - 17 = (n - 1)(-3)

⇒ -57 = (n - 1)(-3)

⇒ $\dfrac{-57}{-3}$ = (n - 1)

⇒ 19 = n - 1

⇒ n = 19 + 1

⇒ n = 20.

The A.P. has 20 terms.

The 10th term from the end is the (n - 10 + 1)th term from the beginning

= 20 - 10 + 1 = 11th term from beginning.

⇒ a_n = a + (n - 1)d

⇒ a₁₁ = 17 + (11 - 1)(-3)

= 17 + 10(-3)

= 17 - 30

= -13.

Assertion (A) is false.

The standard and correct formula for finding the nth term of an Arithmetic Progression.

a_n = a + (n - 1)d

Reason (R) is true.

A is false, R is true

Hence, option 4 is the correct option.

Assertion (A): For an A.P., T₂₂ = 149 and d = 7. Then S₂₂ is 1661.

Reason (R): The sum of first n terms of an A.P. is given by S_n = $\dfrac{n}{2}[2a − (n − 1)d]$.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Given,

a₂₂ = 149

n = 22

d = 7.

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₂₂ = a + (22 - 1)7

⇒ 149 = a + (21)7

⇒ 149 = a + 147

⇒ 149 - 147 = a

⇒ a = 2.

We know that,

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

⇒ S₂₂ = $\dfrac{22}{2}$ (2 + 149)

= 11 × (151)

= 1661.

Assertion (A) is true.

The standard and correct formula for the sum of the first n terms of an A.P.

S_n = $\dfrac{n}{2}$ [2a + (n − 1)d]

Reason (R) is false.

A is true, R is false

Hence, option 3 is the correct option.

Assertion (A): If the sum of first n terms of an A.P. is given S_n = 2n² − n, then its nth term is 4n - 3.

Reason (R): The nth term (t_n) of an A.P. from the end is l - (n - 1)d.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

We know that,

T_n = S_n - S_{n - 1}

Given,

S_n = 2n² - n

S_{n - 1} = 2(n - 1)² - (n - 1)

= 2(n² - 2n + 1) - n + 1

= 2n² - 4n + 2 - n + 1

= 2n² - 5n + 3.

T_n = 2n² - n - (2n² - 5n + 3)

= 2n² - n - 2n² + 5n - 3

= 4n - 3.

Assertion (A) is true.

The nth term (t_n) of an A.P. from the end is l - (n - 1)d.

Reason (R) is true.

Both A and R are true, but R is not the correct explanation of A.

Hence, option 2 is the correct option.