L e t a 1 , a 2 , a 3 , . . , a n b e n p o s i t i v e c o n s e c u t i v e t e r m s o f a n a r i t h m e t i c p r o g r e s s i o n . I f d 0 i s i t s c o m m o n d i f f e r e n c e , t h e n lim n d n

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Solution

Q.
Let $a_{1}, a_{2}, a_{3}, . ., a_{n}$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference, then

$\lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}) i s$ [2023]

1 0

2 $\sqrt{d}$

3 1

4 $\frac{1}{\sqrt{d}}$

Ans.
(3)

$W e h a v e, a_{1}, a_{2}, . . . ., a_{n} a r e i n A . P .$

$∴ a_{2} - a_{1} = a_{3} - a_{2} = . . . . . . = a_{n} - a_{n - 1} = d$

$N o w, \lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}})$

$= \lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{a_{2} - a_{1}} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{a_{3} - a_{2}} + . . . + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{a_{n} - a_{n - 1}})$

$= \lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{d} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{d} + . . . + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{d})$

$= \lim_{n \to \infty} \frac{1}{\sqrt{n d}} (\sqrt{a_{n}} - \sqrt{a_{1}})$

$= \lim_{n \to \infty} \frac{1}{\sqrt{n d}} (\frac{a_{n} - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}}) = \lim_{n \to \infty} \frac{1}{\sqrt{n d}} \frac{(n - 1) d}{\sqrt{a_{n}} + \sqrt{a_{1}}}$

$= \lim_{n \to \infty} \frac{1}{\sqrt{n d}} \frac{n (1 - \frac{1}{n}) d}{\sqrt{n} (\sqrt{\frac{a_{1}}{n} + (1 - \frac{1}{n})} d + \sqrt{\frac{a_{1}}{n}})}$

$= \lim_{n \to \infty} \frac{(1 - \frac{1}{n}) d}{\sqrt{d} (\sqrt{\frac{a_{1}}{n} + d - \frac{d}{n}}) + \sqrt{\frac{a_{1}}{n}}} = \frac{d}{\sqrt{d} \cdot \sqrt{d}} = 1$

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