JEE Main 2025 · 24 January · morning · Single correct · +4 / −1

For some n ≠ 10, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of (1+x)ⁿ⁺⁴ be in A.P. Then the largest coefficient in the expansion of (1+x)ⁿ⁺⁴ is:

  1. (a) 10
  2. (b) 35
  3. (c) 70
  4. (d) 20
Answer and solution

Answer (b) 35

Solution

Maths sir · 4 steps

Put m = n+4, turn the A.P. condition into a quadratic in m, and the condition n ≠ 10 picks the root.

  1. Name the coefficients. Let m = n + 4. The (r+1)th term of (1+x)ᵐ has coefficient C(m, r), so the 5th, 6th and 7th terms have C(m, 4), C(m, 5), C(m, 6).

    2C(m, 5) = C(m, 4) + C(m, 6) ⇒ 2 = 5/(m-4) + (m-5)/6

  2. Solve for m. Divide the A.P. condition by C(m, 5) using C(m, 4)/C(m, 5) = 5/(m-4) and C(m, 6)/C(m, 5) = (m-5)/6. Multiply through by 6(m-4): 12(m-4) = 30 + (m-5)(m-4), which gives m² - 21m + 98 = 0, so (m-7)(m-14) = 0.
  3. Use the condition. m = 14 means n = 10, which is not allowed. So m = 7 and n = 3. Check: C(7, 4), C(7, 5), C(7, 6) = 35, 21, 7, an A.P. with common difference -14.
  4. So. In (1+x)⁷ the exponent is odd, so the two middle terms share the largest coefficient: C(7, 3) = C(7, 4) = 35. The answer is (b).

The idea to keep. For coefficients in A.P., divide everything by the middle binomial coefficient so that only simple ratios remain. Then the largest coefficient of (1+x)ᵐ is the middle one, C(m, ⌊ m/2 ⌋).

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2025.