If the sum of first three natural numbers from a group of six consecutive natural numbers is 36, then the sum of last three natural numbers will be :

Question

Accepted Answer

To solve the problem, we need to determine the sum of the last three natural numbers in a group of six consecutive natural numbers, given that the sum of the first three is 36.

Let's denote the six consecutive natural numbers as $ n, n+1, n+2, n+3, n+4, n+5 $.

1. **Sum of the first three numbers:**
   $$
   n + (n+1) + (n+2) = 3n + 3
   $$
   We are given that this sum is 36:
   $$
   3n + 3 = 36
   $$
   Solving for $ n $:
   $$
   3n = 36 - 3 = 33
   $$
   $$
   n = 11
   $$

2. **Identifying the six numbers:**
   With $ n = 11 $, the six consecutive numbers are:
   $$
   11, 12, 13, 14, 15, 16
   $$

3. **Sum of the last three numbers:**
   The last three numbers are $ 14, 15, $ and $ 16 $.
   $$
   14 + 15 + 16 = 45
   $$

Thus, the sum of the last three natural numbers is 45, which corresponds to Option 1.

**Explanation for why other options are incorrect:**

- **Option 2 (52):** This would imply a different set of numbers or incorrect calculation, which does not align with the given conditions.
- **Option 3 (56):** Similarly, this sum does not match the sum of the last three numbers in the sequence derived from the given conditions.
- **Option 4 (55):** This is also incorrect as it does not correspond to the sum of the last three numbers calculated from the sequence.

Therefore, based on the logical steps and calculations, Option 1 is indeed the correct answer.