Let Sₙ denote the sum of first n-terms of an arithmetic progression. If S₁₀ = 530, S₅ = 140, then S₂₀ - S₆ is equal to:

Solution

Maths sir · 4 steps

Two sums give two equations in a and d. Find both, then the rest is plugging in.

1. Set up equations. Use Sₙ = n/2(2a + (n-1)d). From S₁₀ = 5(2a + 9d) = 530 you get 2a + 9d = 106. From S₅ = 5/2(2a + 4d) = 140 you get 2a + 4d = 56.
2. Find a and d. Subtracting the two equations gives 5d = 50, so d = 10. Then 2a = 56 - 40 = 16, so a = 8.
3. Compute both sums. S₂₀ = 10(16 + 19 × 10) = 10 × 206 = 2060. S₆ = 3(16 + 5 × 10) = 3 × 66 = 198.

   S₂₀ - S₆ = 2060 - 198 = 1862

4. So. The answer is 1862, option (a). As a check, S₂₀ - S₆ is the sum of terms 7 to 20. That is 14 terms from a₇ = 68 to a₂₀ = 198, so 14/2(68 + 198) = 7 × 266 = 1862.

The idea to keep. Any two values of Sₙ in an AP fix a and d. Also remember that Sₘ - Sₖ is the sum of terms k+1 to m, which gives you a quick cross-check.

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