If a 1 - a 2 - a 3 - a n are in harmonical progression- show that a n a n-1 - 1 2- 1 2- 1 2- - 1 2- a 2 a 1-

⇒ By definition of H.P. we have; 1 a _ n + 1 a _ n 2 = b _ 3 a _ n 1 ∴ a _ n 1 a _ n =2 a _ n 1 a _ n 2 a _ n a _ n 1 1 2 a _ n ...

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Standard XII

Mathematics

Vn Method

If a 1 , a ...

Question

If $a_{1}, a_{2}, a_{3}, \dots a_{n}$ are in harmonical progression, show that
$\frac{a_{n}}{a_{n - 1}} = \frac{1}{2 -} \frac{1}{2 -} \frac{1}{2 -} \dots \frac{1}{2 -} \frac{a_{2}}{a_{1}}$ .

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Solution

$\Rightarrow$ By definition of H.P. we have;-
$\frac{1}{a_{n}} + \frac{1}{a_{n - 2}} = \frac{b_{3}}{a_{n - 1}}$
$∴ \frac{a_{n - 1}}{a_{n}} = 2 - \frac{a_{n - 1}}{a_{n - 2}}$
$\frac{a_{n}}{a_{n - 1}} \frac{1}{2 - \frac{a_{n - 1}}{a_{n - 2}}}; \frac{a_{n - 1}}{a_{n - 2}}; \frac{1}{2 - \frac{a_{n - 2}}{a_{n - 3}}}$
and so on
$\frac{a_{3}}{a_{2}} = \frac{1}{2 - \frac{a_{2}}{a_{1}}}$
$∴ \frac{a_{n}}{a_{n - 1}} = \frac{1}{2 -} \frac{1}{2 -} \frac{1}{2 -} . . . . . \frac{1}{2 -} \frac{a_{2}}{a_{1}}$

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