Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11ᵗʰ term is :

Solution

Maths sir · 5 steps

The first three terms give you a + d = 18. Now the sum bound pins d to a single integer.

1. Use the first condition. The first three terms are a, a + d, a + 2d, so their sum is 3a + 3d = 54. This gives a + d = 18, so a = 18 - d.
2. Write the twenty-term sum. S₂₀ = 20/2(2a + 19d) = 10(2a + 19d). Put a = 18 - d into it.

   S₂₀ = 10(36 - 2d + 19d) = 360 + 170d

3. Trap the difference. 1600 < 360 + 170d < 1800 gives 1240 < 170d < 1440, so 7.29 < d < 8.47. The terms are integers, so d is an integer, and the only choice is d = 8.
4. Find the first term. a = 18 - 8 = 10, which is a positive integer. As a check, S₂₀ = 360 + 1360 = 1720, which lies between 1600 and 1800.
5. So. T₁₁ = a + 10d = 10 + 80 = 90, which is option (b).

The idea to keep. When an A.P. question gives a sum as a range, write that sum in terms of one unknown. Then use the fact that the terms are integers to pick out the single value that fits.

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