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

Question

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

Answer 1

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 $2^3=8$, and $8\mathrm{mod}3=2$.
• The cube of $5$ is $5^3=125$, and $125\mathrm{mod}3=2$.
• The cube of $8$ is $8^3=512$, and $512\mathrm{mod}3=2$.

In fact, for any number of the form $3n+2$ (let's denote it by $k$), we have: $$k=3n+2$$ $$k^3=(3n+2{)}^3=27n^3+3(3n^2)(2)+3(3n)(2^2)+2^3=27n^3+54n^2+36n+8$$ When we take $k^3\mathrm{mod}3$, all terms except $8$ are divisible by $3$: $$k^3\equiv 8\mathrm{mod}3\equiv 2\mathrm{mod}3$$

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.