In an arithmetic progression, if S₄₀=1030 and S₁₂=57, then S₃₀-S₁₀ is equal to :

Solution

Maths sir · 4 steps

S₃₀-S₁₀ is just terms 11 to 30, and their end-pair sum equals a₁+a₄₀. So it is half of S₄₀.

1. What is being asked. S₃₀-S₁₀ removes the first 10 terms from the first 30. What is left is a₁₁+a₁₂+…+a₃₀, which is 20 terms of the same AP.
2. Sum of these terms. The sum of an AP block is (number of terms) × (first + last) ÷ 2. Here the ends are a₁₁=a+10d and a₃₀=a+29d, so a₁₁+a₃₀=2a+39d. That is exactly a₁+a₄₀.

   S₃₀-S₁₀=20/2(2a+39d)=1/2·40/2(2a+39d)=S₄₀/2

3. Quick check. From S₄₀=1030 we get 2a+39d=51.5. From S₁₂=57 we get 2a+11d=9.5. Subtracting gives d=1.5 and a=-3.5. Then S₃₀=547.5 and S₁₀=32.5, and their difference is 515.
4. So. S₃₀-S₁₀=1030/2= 515, which is option (d).

The idea to keep. In an AP, terms the same distance from the ends of a block add to the same value. Before solving for a and d, check whether the block you need pairs up like a sum you already know.

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