Correct answer is B. 2S
Let the common difference and number of terms of AP be d and n respectively.
Last term of AP = an = l (given)
l = a + (n–1) d
D = \(\frac{(a + l)(a-l)}{2s - (L + a)}\)
Comparing this with \(\frac{1^2 -a^2}{K - (1 +a)}\)
We get k =2S