Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ___________.

Solution

Maths sir · 2 steps

Last term minus first term is 99, so the number of gaps times d must equal 99. Just read off divisors.

1. Set up the condition. With n terms the last term is 100 + (n-1)d = 199. So (n-1)d = 99, which means n-1 must be a positive divisor of 99 and d = 99/(n-1).

   (n-1) d = 99, 3 ≤ n ≤ 33 ⇒ 2 ≤ n-1 ≤ 32

   n-1 (divisor of 99)	n	Allowed?	d
   1	2	no, fewer than 3 terms	99
   3	4	yes	33
   9	10	yes	11
   11	12	yes	9
   33	34	no, more than 33 terms	3
   99	100	no, more than 33 terms	1

2. So. The allowed common differences are 33, 11 and 9, and their sum is 33 + 11 + 9 = 53.

The common slip: 56. Putting the range 3 to 33 on n-1 instead of on n lets in n-1 = 33 with d = 3, which gives 56.

The idea to keep. When the first and last terms are fixed, (n-1)d is fixed. Turn the AP into a divisor-counting problem, and put the limits on n-1, not on n.

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