JEE Main 2023 · 1 February · evening · Numerical answer · +4 / −0
The sum of the common terms of the following three arithmetic progressions.
3,7,11,15, … ., 399,
2,5,8,11, … ., 359 and
2,7,12,17, … ., 197,
is equal to _____________.
Answer and solution
Answer 321
Solution
Maths sir · 4 steps
Common terms step by LCM(4, 3, 5) = 60. Find the first one, then stop at the smallest last term.
- Common difference. The differences are 4, 3 and 5. Consecutive terms common to all three progressions are LCM(4, 3, 5) = 60 apart, so the common terms form an AP with difference 60.
- First common term. The second and third APs both start at 2, so their common terms go up by LCM(3, 5) = 15: 2, 17, 32, 47, … The first AP holds only numbers that leave remainder 3 when divided by 4. In this list, the first such number is 47 = 4 × 11 + 3.
- Upper limit. A common term has to lie in all three lists, so it cannot go past the smallest last term, which is 197. The common terms are 47, 107, 167. The next one, 227, is already beyond 197.
47 + 107 + 167 = 321
- So. The sum of the common terms is 321.
The common slip: 1182. Using 359 as the upper limit instead of 197 adds 227, 287 and 347, giving 1182. The third AP ends at 197, so no common term can be larger.
The idea to keep. Common terms of several APs form a new AP. Its difference is the LCM of the given differences, it starts at the first common term, and it stops at the smallest last term.
Answer from the NTA final answer key. Solution by PrepWiser.