The sum of all those terms, of the arithmetic progression 3, 8, 13, ...., 373, which are not divisible by 3, is equal to ____________.

Solution

Maths sir · 3 steps

Do not hunt for the unwanted terms one by one: add everything, then subtract the multiples of 3.

1. Whole progression. Here a = 3 and d = 5. From 3 + 5(n-1) = 373 we get n = 75 terms. Their sum is 75/2(3 + 373) = 75 × 188 = 14100.
2. Find multiples of 3. The first term, 3, is a multiple of 3. Multiples of 3 repeat every lcm(3, 5) = 15, so they form a new AP: 3, 18, 33, … The largest one up to 373 is 363 = 3 + 15 × 24, which gives 25 such terms.

   S_(div by 3) = 25/2(3 + 363) = 25 × 183 = 4575

3. So. The sum of the terms not divisible by 3 is 14100 - 4575 = 9525.

The common slip: 4575. Stopping after the second sum and reporting the total of the multiples of 3 gives 4575, not 9525. Subtract that from the full sum of 14100.

The idea to keep. To sum the terms of an AP that are not multiples of k, find the multiples as their own AP with common difference lcm(d, k), then subtract that sum from the total.

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