If 1/(20-a)(40-a)+1/(40-a)(60-a)+…+1/(180-a)(200-a)=1/256, then the maximum value of a is :

Solution

Maths sir · 4 steps

The factors always differ by 20, so split each term into two fractions and let the middle cancel.

1. Split each term. In every term the two factors differ by 20. For example, (40-a)-(20-a)=20. So each term can be written as 1/uv=1/20(1/u-1/v).
2. Telescope. When you add the terms, every middle fraction cancels. Only the first and the last fractions are left: 1/20(1/(20-a)-1/(200-a))=1/20·180/(20-a)(200-a)=9/(20-a)(200-a).

   9/(20-a)(200-a)=1/256 ⇒ (20-a)(200-a)=2304

3. Solve the quadratic. Expanding gives a²-220a+4000=2304, that is a²-220a+1696=0. The discriminant is 48400-6784=41616=204², so a=(220±204)/2=212 or 8.
4. So. The larger root is a=212. Check: (20-212)(200-212)=(-192)(-12)=2304, and no factor in the sum becomes zero. The maximum value of a is 212, option (c).

The idea to keep. When each term is 1/uv with v-u=d fixed, write it as 1/d(1/u-1/v). Only the first and last pieces survive the sum.

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