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Question Number 142865 by EnterUsername last updated on 06/Jun/21

Let a_1 , a_2 , a_3 , ... be an arithmethic progression of positive real numbers. Then (1/( (√a_1 )+(√a_2 )))+(1/( (√a_2 )+(√a_3 )))+∙∙∙+(1/( (√a_(n−1) )+(√a_n )))= (A) ((n+1)/( (√a_1 )+(√a_n ))) (B) ((n−1)/( (√a_1 )+(√a_n ))) (C) (n/( (√a_1 )+(√a_n ))) (D) (n/( (√a_n )−(√a_1 )))

$$\mathrm{Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\:...\:\mathrm{be}\:\mathrm{an}\:\mathrm{arithmethic}\:\mathrm{progression}\:\mathrm{of} \\ $$$$\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}.\:\mathrm{Then} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{{a}_{\mathrm{2}} }+\sqrt{{a}_{\mathrm{3}} }}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\:\sqrt{{a}_{{n}−\mathrm{1}} }+\sqrt{{a}_{{n}} }}= \\ $$$$\left(\mathrm{A}\right)\:\frac{{n}+\mathrm{1}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{{n}} }}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{{n}−\mathrm{1}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{{n}} }} \\ $$$$\left(\mathrm{C}\right)\:\frac{{n}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{{n}} }}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{{n}}{\:\sqrt{{a}_{{n}} }−\sqrt{{a}_{\mathrm{1}} }} \\ $$

Commented by som(math1967) last updated on 07/Jun/21

(B)((n−1)/( (√a_1 )+(√a_n ))) □

$$\left({B}\right)\frac{{n}−\mathrm{1}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{{n}} }}\:\:\square \\ $$

Answered by mr W last updated on 06/Jun/21

(1/( (√a_1 )+(√a_2 )))+(1/( (√a_2 )+(√a_3 )))+∙∙∙+(1/( (√a_(n−1) )+(√a_n ))) =(((√a_2 )−(√a_1 ))/( a_2 −a_1 ))+(((√a_3 )−(√a_2 ))/( a_3 −a_2 ))+∙∙∙+(((√a_n )−(√a_(b−1) ))/( a_n −a_(n−1) )) =(((√a_n )−(√a_1 ))/d) =((((√a_n )−(√a_1 ))(n−1))/(a_n −a_1 )) =((n−1)/( (√a_n )+(√a_1 ))) ⇒answer (B)

$$\frac{\mathrm{1}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{{a}_{\mathrm{2}} }+\sqrt{{a}_{\mathrm{3}} }}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\:\sqrt{{a}_{{n}−\mathrm{1}} }+\sqrt{{a}_{{n}} }} \\ $$$$=\frac{\sqrt{{a}_{\mathrm{2}} }−\sqrt{{a}_{\mathrm{1}} }}{\:{a}_{\mathrm{2}} −{a}_{\mathrm{1}} }+\frac{\sqrt{{a}_{\mathrm{3}} }−\sqrt{{a}_{\mathrm{2}} }}{\:{a}_{\mathrm{3}} −{a}_{\mathrm{2}} }+\centerdot\centerdot\centerdot+\frac{\sqrt{{a}_{{n}} }−\sqrt{{a}_{{b}−\mathrm{1}} }}{\:{a}_{{n}} −{a}_{{n}−\mathrm{1}} } \\ $$$$=\frac{\sqrt{{a}_{{n}} }−\sqrt{{a}_{\mathrm{1}} }}{{d}} \\ $$$$=\frac{\left(\sqrt{{a}_{{n}} }−\sqrt{{a}_{\mathrm{1}} }\right)\left({n}−\mathrm{1}\right)}{{a}_{{n}} −{a}_{\mathrm{1}} } \\ $$$$=\frac{{n}−\mathrm{1}}{\:\sqrt{{a}_{{n}} }+\sqrt{{a}_{\mathrm{1}} }} \\ $$$$\Rightarrow{answer}\:\left({B}\right) \\ $$

Commented by EnterUsername last updated on 06/Jun/21

a_n =a_1 +(n−1)d. Thank you, Sir

$${a}_{{n}} ={a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}.\: \\ $$$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{Sir} \\ $$

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