The sum of three consecutive multiples of 3 is 99. These multiples are

Method 1 (let the middle multiple be the variable):

Let the three consecutive multiples of 3 be 3n, 3(n+1), and 3(n+2).

Sum = 3n + 3(n+1) + 3(n+2) = 9n + 9

Given sum = 99, so 9n + 9 = 99 → 9n = 90 → n = 10

So the multiples are: 3×10 = 30, 3×11 = 33, 3×12 = 36.

Method 2 (assume the multiples as x, x+3, x+6):

Since the numbers are consecutive multiples of 3, they differ by 3, so we can write them as x, x+3, and x+6, where x is itself a multiple of 3.

Sum = x + (x+3) + (x+6) = 3x + 9 = 99 → 3x = 90 → x = 30.

So the multiples are 30, 33, 36 — the same answer.

Hence, the numbers are 30, 33, 36.

हिन्दी हल:

विधि 1: माना तीन क्रमागत गुणज 3n, 3(n+1), और 3(n+2) हैं। योग = 9n + 9 = 99 → n = 10. अतः गुणज 30, 33, 36 हैं।

विधि 2: संख्याओं को x, x+3, x+6 मान सकते हैं। योग = 3x + 9 = 99 → x = 30. अतः संख्याएँ 30, 33, 36 हैं — वही उत्तर।