Let Sₙ denote the sum of the first n terms of an arithmetic progression. If S₁₀=390 and the ratio of the tenth and the fifth terms is 15: 7, then S₁₅-S₅ is equal to :

Solution

Maths sir · 5 steps

Turn both conditions into equations in a and d, solve them, and the rest is plain adding.

1. Use the sum. S₁₀ = 10/2(2a + 9d) = 390, so 2a + 9d = 78.
2. Use the ratio. (a + 9d)/(a + 4d) = 15/7 gives 7a + 63d = 15a + 60d, so 8a = 3d.
3. Find a and d. Put a = 3d/8 into 2a + 9d = 78. This gives 3d/4 + 9d = 39d/4 = 78, so d = 8 and a = 3.

   S₁₅ - S₅ = a₆ + a₇ + … + a₁₅ = 10/2(a₆ + a₁₅)

4. Add the terms. a₆ = 3 + 5 × 8 = 43 and a₁₅ = 3 + 14 × 8 = 115. So the sum is 5 × 158 = 790. As a check, S₁₅ = 885 and S₅ = 95, and 885 - 95 = 790.
5. So. S₁₅ - S₅ = 790, which is option (c).

The idea to keep. Any AP question with two conditions is really two equations in a and d. And Sₘ - Sₙ is just the terms from aₙ₊₁ to aₘ, which is again an AP sum.

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