Let 3,7,11,15, …, 403 and 2,5,8,11, …, 404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to ___________.

Solution

Maths sir · 4 steps

The common terms form their own AP: find its first term, its step and its last term, then add.

1. First common term. List a few terms of each: 3, 7, 11, 15, … and 2, 5, 8, 11, …. The first number in both lists is 11.
2. Common difference. After 11, the next common term comes when both steps line up again. That happens after LCM(4, 3) = 12, so the common terms are 11, 23, 35, ….
3. Last common term. A common term cannot go past the smaller last term, 403. Solve 11 + 12k ≤ 403 to get k ≤ 32.67, so k = 32 and the last common term is 11 + 384 = 395. Check: 395 = 4 × 99 - 1 and 395 = 3 × 132 - 1, so it lies in both. That gives k = 0, 1, …, 32, which is 33 terms.

   S = 33/2(11 + 395) = 33 × 203

4. So. The sum of the common terms is 33 × 203 = 6699.

The common slip: 6304. Counting the terms as (395 - 11)/12 = 32 and forgetting to add 1 gives a sum that stops at 383, which is 6304.

The idea to keep. Common terms of two APs form a new AP. Its first term is the first shared number, its difference is the LCM of the two differences, and it stops at or before the smaller last term.

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