Assertion (A): a, b, c are in AP if and only if 2b = a + c.

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Q.
Assertion (A): a, b, c are in AP if and only if 2b = a + c.

Reason (R): The sum of first n odd natural numbers is n².

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.

Ans.
(2)

Since a, b, c are in AP.

$∴ b - a = c - b \Rightarrow 2 b = a + c A l s o, s u m o f f i r s t n o d d n a t u r a l n u m b e r s = 1 + 3 + 5 + . . . w h i c h a r e i n A P w i t h a = 1, d = 3 - 1 = 2 ∴ S ₙ = n / 2 (2 a + (n - 1) d) = n / 2 (2 \times 1 + (n - 1) \times 2) = n (1 + n - 1) = n ²$

Solution

Q.
Assertion (A): a, b, c are in AP if and only if 2b = a + c.

Reason (R): The sum of first n odd natural numbers is n².

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.

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Solution

Q. Assertion (A): a, b, c are in AP if and only if 2b = a + c. Reason (R): The sum of first n odd natural numbers is n².

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.

Ans. (2) Since a, b, c are in AP. ∴ b – a = c – b ⇒ 2b = a + cAlso, sum of first n odd natural numbers= 1 + 3 + 5 + ... which are in AP with a = 1, d = 3 – 1 = 2∴ Sₙ = n/2 (2a + (n – 1)d) = n/2 (2×1 + (n – 1)×2) = n(1 + n – 1) = n²

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Q.
Assertion (A): a, b, c are in AP if and only if 2b = a + c.

Reason (R): The sum of first n odd natural numbers is n².