Question

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.

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 \(8 \mod 3 = 2\).
- The cube of \(5\) is \(5^3 = 125\), and \(125 \mod 3 = 2\).
- The cube of \(8\) is \(8^3 = 512\), and \(512 \mod 3 = 2\).

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 \(k^3 \mod 3\), all terms except \(8\) are divisible by \(3\):
\[
k^3 \equiv 8 \mod 3 \equiv 2 \mod 3
\]

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.