Arithmetic Sequence 24485B

1. **Stating the problem:** We want to determine which of the given sequences $\left(a_n\right)$ is an arithmetic sequence. 2. **Recall the definition:** A sequence $\left(a_n\right)$ is arithmetic if the difference between consecutive terms is constant, i.e., $a_{n+1} - a_n = d$ for some constant $d$. --- ### Sequence ក: $a_n = \frac{n^2 - 1}{n}$ 3. Simplify $a_n$: $$a_n = \frac{n^2 - 1}{n} = \frac{(n-1)(n+1)}{n} = n + \frac{1}{n} - 1$$ 4. Calculate the difference: $$a_{n+1} - a_n = \left((n+1) + \frac{1}{n+1} - 1\right) - \left(n + \frac{1}{n} - 1\right) = (n+1 - 1) - (n - 1) + \frac{1}{n+1} - \frac{1}{n} = 1 + \frac{1}{n+1} - \frac{1}{n}$$ 5. Simplify the fractional part: $$\frac{1}{n+1} - \frac{1}{n} = \frac{n - (n+1)}{n(n+1)} = \frac{-1}{n(n+1)}$$ 6. So, $$a_{n+1} - a_n = 1 - \frac{1}{n(n+1)}$$ 7. Since this depends on $n$, the difference is not constant. Therefore, sequence ក is **not arithmetic**. --- ### Sequence ខ: $a_n = \frac{1}{\sqrt{n^2 + 1} - n}$ 8. Rationalize the denominator: $$a_n = \frac{1}{\sqrt{n^2 + 1} - n} \times \frac{\sqrt{n^2 + 1} + n}{\sqrt{n^2 + 1} + n} = \frac{\sqrt{n^2 + 1} + n}{(\sqrt{n^2 + 1})^2 - n^2} = \frac{\sqrt{n^2 + 1} + n}{1} = \sqrt{n^2 + 1} + n$$ 9. Calculate the difference: $$a_{n+1} - a_n = \left(\sqrt{(n+1)^2 + 1} + (n+1)\right) - \left(\sqrt{n^2 + 1} + n\right) = \sqrt{n^2 + 2n + 2} - \sqrt{n^2 + 1} + 1$$ 10. This difference depends on $n$ and is not constant, so sequence ខ is **not arithmetic**. --- ### Sequence គ: $a_n = \frac{n + 1}{n + 3}$ 11. Calculate the difference: $$a_{n+1} - a_n = \frac{(n+1) + 1}{(n+1) + 3} - \frac{n + 1}{n + 3} = \frac{n + 2}{n + 4} - \frac{n + 1}{n + 3}$$ 12. Find common denominator and subtract: $$a_{n+1} - a_n = \frac{(n + 2)(n + 3) - (n + 1)(n + 4)}{(n + 4)(n + 3)}$$ 13. Expand numerator: $$(n + 2)(n + 3) = n^2 + 3n + 2n + 6 = n^2 + 5n + 6$$ $$(n + 1)(n + 4) = n^2 + 4n + n + 4 = n^2 + 5n + 4$$ 14. Subtract numerator: $$n^2 + 5n + 6 - (n^2 + 5n + 4) = 6 - 4 = 2$$ 15. So, $$a_{n+1} - a_n = \frac{2}{(n + 4)(n + 3)}$$ 16. Since denominator depends on $n$, difference is not constant. Sequence គ is **not arithmetic**. --- ### **Conclusion:** None of the sequences ក, ខ, or គ have a constant difference between consecutive terms, so **none of them is an arithmetic sequence**.