If-a-1-a-2-a-n-be-an-arithmetic-progression-then-show-that-1-a-1-a-n-1-a-2-a-n-1-1-a-3-a-n-2-1-a-n-a-1-2-a-1-a-n-1-a-1-1-a-2-

Question Number 4704 by lakshaysethi039 last updated on 25/Feb/16

I f a_{1}, a_{2}, \dots \dots \dots a_{n} b e a n a r i t h m e t i c p r o g r e s s i o n,

t h e n s h o w t h a t

\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} + \dots \dots \dots \dots . + \frac{1}{a_{n} a_{1}}

= \frac{2}{a_{1} + a_{n}} (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \dots \dots + \frac{1}{a_{n}})

Commented by Yozzii last updated on 25/Feb/16

I n t e r e s t i n g p r o p e r t y o f A . P s o f

n o n - z e r o t e r m s . I^{'} v e n e v e r s e e n t h i s

b e f o r e .

Answered by Yozzii last updated on 25/Feb/16

L e t S_{n} = \underset{i = 1}{\sum^{n}} a_{i}^{- 1} a n d J = \frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} \dots + \frac{1}{a_{n} a_{1}} .

F i r s t c a l c u l a t e s u m s t h a t y i e l d t h e

d e n o m i n a t o r s o f t e r m s o f J i n t e r m s o f a_{i} .

(1) a_{1}^{- 1} + a_{n}^{- 1} = \frac{a_{1} + a_{n}}{a_{1} a_{n}}

(2) a_{2}^{- 1} + a_{n - 1}^{- 1} = \frac{a_{2} + a_{n - 1}}{a_{2} a_{n - 1}}

(3) a_{3}^{- 1} + a_{n - 2}^{- 1} = \frac{a_{3} + a_{n - 2}}{a_{3} a_{n - 2}}

(4) a_{4}^{- 1} + a_{n - 3}^{- 1} = \frac{a_{4} + a_{n - 3}}{a_{4} a_{n - 3}}

⋮

(n) a_{n}^{- 1} + a_{1}^{- 1} = \frac{a_{n} + a_{1}}{a_{1} a_{n}}

A d d i n g l i n e s (1) \to (n) w e s e e t h a t

a l l a_{i}^{- 1} a r e a d d e d t w i c e . H e n c e, w e m a y

w r i t e

2 (\underset{i = 1}{\sum^{n}} a_{i}^{- 1}) = \frac{a_{1} + a_{n}}{a_{1} a_{n}} + \frac{a_{2} + a_{n - 1}}{a_{2} a_{n - 1}} + \frac{a_{3} + a_{n - 2}}{a_{3} a_{n - 2}} + \frac{a_{4} + a_{n - 3}}{a_{4} a_{n - 3}} + \dots + \frac{a_{n} + a_{1}}{a_{1} a_{n}} . (*)

W e a r e g i v e n t h a t {a_{i}}_{i = 1}^{n} i s a n

a r i t h m e t i c p r o g r e s s i o n . T h e r e f o r e,

a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots = a_{n - 2} - a_{n - 3} = a_{n - 1} - a_{n - 2} = a_{n} - a_{n - 1} .

U s i n g t h e s e e q u a t i o n s w e o b t a i n

a_{2} - a_{1} = a_{n} - a_{n - 1}, o r a_{1} + a_{n} = a_{2} + a_{n - 1} .

S i m i l a r l y w e h a v e a_{n - 1} - a_{n - 2} = a_{3} - a_{2}

s o t h a t a_{n - 2} + a_{3} = a_{n - 1} + a_{2} = a_{1} + a_{n} .

A l s o, a_{n - 2} - a_{n - 3} = a_{4} - a_{3}

\Rightarrow a_{4} + a_{n - 3} = a_{3} + a_{n - 2} = a_{1} + a_{n} .

U p o n r e p e a t e d u s e o f t h i s p r o c e s s w e

o b t a i n t h e n u m e r a t o r s o f t h e t e r m s o n

t h e r . h . s o f (*) a l l b e i n g e q u a l t o a_{1} + a_{n} .

∴ (a_{1} + a_{n}) {\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} + \frac{1}{a_{4} a_{n - 3}} + \dots + \frac{1}{a_{1} a_{n}}} = 2 S_{n}

o r (a_{1} + a_{n}) J = 2 S_{n}

\Rightarrow J = \frac{2}{a_{1} + a_{n}} S_{n}

T h i s i s v a l i d i f f a_{i} \neq 0 f o r 1 ⩽ i ⩽ n .

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