b,a,d,c,...,x,y (they are n numbers and n is Subset of set of natural numbers) are an Arithmetic sequence and all of them are Positive so prove this phrase

(1/(√b+√a))+(1/(√d+√c))+...+(1/(√x+√y))=(n-1)/(√b+√y)

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b,a,d,c,...,x,y (they are n numbers and n is Subset of set of natural numbers) are an Arithmetic sequence and all of them are Positive so prove this phrase

(1/(√b+√a))+(1/(√d+√c))+...+(1/(√x+√y))=(n-1)/(√b+√y)

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## Answers (1)

Let us take a, b, c, ..., x, y, z as

a₁, a₂, a₃, ...., a_n respectively.

Then , rationalizing each denominator,

... LHS

= [ 1/ (√a₁ + √a₂ ) ] + [ 1/ (√a₂ + √a₃ ) ] + ... + [ 1/ (√a_(n-1) + √a_n ]

= [ ( √a₂ - √a₁ ) / ( a₂ - a₁ ) ] + [ ( √a₃ - √a₂ ) / ( a₃ - a₂ ) ] + ..... to (n-1) brackets

= [ ( √a₂ - √a₁ ) / ( d ) ] + [ ( √a₃ - √a₂ ) / ( d ) ] + ..... to (n-1) brackets

= [ ( √a₂ - √a₁ ) + ( √a₃ - √a₂ ) + ... + ( √a_n - √a_(n-1) ) ] / d

= ( - √a₁ + √a_n ) / d

= ( √a_n - √a₁ ) / d ...................................... Now rationalize Numerator

= [ (a_n) - (a₁) ] / [ d( √a_n) + √a₁ ) ]

= [ ( a + (n-1)d) - a ] / [ d( √a + √z ) ]

= (n-1)d / [ d( √a + √z ) ]

= ( n - 1 ) / ( √a + √z )

= RHS ........................................… Q.E.D.

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