48. Assertion: 4,7,10,... are the natural numbers of the form 3n + 1 where n =1,2,3…

Reason: The cube of natural number of the form 3n + 2 when divided by 3 leaves remainder 2.

1 answer

To assess the assertion and reason provided, let’s break them down one by one:

Assertion: 4, 7, 10,... are the natural numbers of the form 3n+1 where n=1,2,3,....

  • The numbers of the form 3n+1 can be calculated for n=1,2,3,...:
    • For n=1: 3(1)+1=4
    • For n=2: 3(2)+1=7
    • For n=3: 3(3)+1=10

So, the assertion is true as the numbers 4, 7, 10 indeed follow the form 3n+1.

Reason: The cube of a natural number of the form 3n+2 when divided by 3 leaves a remainder of 2.

  • Let’s explore 3n+2:
    • For n=0: 3(0)+2=2
    • For n=1: 3(1)+2=5
    • For n=2: 3(2)+2=8

Now, let’s compute the cubes of these numbers:

  • The cube of 2 is 23=8, and 8mod3=2.
  • The cube of 5 is 53=125, and 125mod3=2.
  • The cube of 8 is 83=512, and 512mod3=2.

In fact, for any number of the form 3n+2 (let's denote it by k), we have: k=3n+2 k3=(3n+2)3=27n3+3(3n2)(2)+3(3n)(22)+23=27n3+54n2+36n+8 When we take k3mod3, all terms except 8 are divisible by 3: k38mod32mod3

So the reason is also true.

Conclusion: Both the assertion and the reason are true. Therefore, the answer would be that both the assertion and the reason are correct, and the reason supports the assertion.