Sequences & Series
Limit of Sum via Telescoping
nta_pyq_2023_apr
Grade 11
Question:
Let $a_1,a_2,\ldots,a_n$ be positive AP terms with $d>0$. Then $\displaystyle\lim_{n\to\infty}\dfrac{\sqrt{d}}{n}\left(\dfrac{1}{\sqrt{a_1}+\sqrt{a_2}}+\cdots+\dfrac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right)$ is
$\dfrac{1}{\sqrt{d}}$
$\sqrt{d}$
$1$
$2$
Step-by-Step Solution
Key Concept: Rationalize: $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$. Sum telescopes to $\frac{\sqrt{a_n}-\sqrt{a_1}}{d}$.
Limit $=1$.
Correct Answer: 3