If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is :

Solution

Maths sir · 4 steps

With n means inserted there are n + 1 gaps, so the common difference is (100 − a)/(n + 1).

1. Set up the AP. The terms a, A₁, A₂, …, Aₙ, 100 form an AP with n + 2 terms, so there are n + 1 equal gaps. That gives d = (100 - a)/(n + 1), the first mean A₁ = a + d and the last mean Aₙ = 100 - d.
2. Use the ratio. Aₙ = 7A₁ gives 100 - d = 7a + 7d, so 7a + 8d = 100. Now put in a = 33 - n, which makes d = (67 + n)/(n + 1).

   7(33 - n) + 8(67 + n)/(n + 1) = 100

3. Clear the fraction. Multiply through by n + 1: (131 - 7n)(n + 1) + 536 + 8n = 0, which becomes 7n² - 132n - 667 = 0. This factorises as (n - 23)(7n + 29) = 0, and n must be a positive integer, so n = 23.
4. So. n = 23, which gives a = 10 and d = 3.75. Check: A₁ = 13.75, Aₙ = 96.25, and 96.25 = 7 × 13.75. The answer is (c).

The idea to keep. Inserting n means between two numbers makes n + 2 terms and n + 1 gaps. Write the first mean as a + d and the last as b − d, then use the given condition.

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