Let Sₙ denote the sum of first n terms of an arithmetic progression. If S₂₀=790 and S₁₀=145, then S₁₅-S₅ is :

Solution

Maths sir · 4 steps

Two sums give two equations in a and d; solve them first, then build S₁₅ and S₅.

1. Sₙ = n/2(2a + (n-1)d)

2. Set up equations. S₂₀ = 10(2a + 19d) = 790 gives 2a + 19d = 79. S₁₀ = 5(2a + 9d) = 145 gives 2a + 9d = 29.
3. Find a and d. Subtracting the two equations gives 10d = 50, so d = 5. Then 2a = 29 - 45 = -16, so a = -8.
4. Compute both sums. S₁₅ = 15/2(-16 + 70) = 15/2 × 54 = 405. S₅ = 5/2(-16 + 20) = 10.
5. So. S₁₅ - S₅ = 405 - 10 = 395, which is option (d). As a check, this is the sum of terms 6 to 15: 10/2(a₆ + a₁₅) = 5(17 + 62) = 395.

The idea to keep. Any two values of Sₙ fix an AP completely: write each as n/2(2a + (n-1)d), solve for a and d, and then any other sum follows.

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