Question

The no. of ways 3 numbers can be selected from the first 2n + 1 natural numbers so that they form an A.P.

• A. n²
• B. n
• C. ⁿC₂
• D. ⁽ⁿ⁺¹⁾C₂

Answer 1

Answer: (A)

n²

Answer 2

Common diff. can vary from 1 to n.

No. of ways with c.d 1 = 2n - 1

No. of ways with c.d 2 = 2n – 3

No. of ways c.d 3 = 2n – 5

No. of ways with c.d n = 1

Total no. of ways = (2n - 1) + (2n - 3) ........ + 1 = n².

Method 2: $b = \frac{a + c}{2}$.

⇒ a, c are both even or both odd.

there n + 1 odd and n even nos.

Total no. of ways = $n + 1C_{2} +^{n}C_{2}$.