Grade 11Algebra
If a1,a2,a3,....,an are in AP,ai>0 for all i,show that 1/(√a1+√a2)+1/(√a2+√a3)+.....+1/(√an-1 + √an)=(n-1)/(√a1+√an)
3 Answers
moumi roy
1/(√a1+√a2)+1/(√a2+√a3)+.....+1/(√an-1 + √an)
(√a2-√a3)/(a2-a3)+...(√an-1-√an)/({an-1}-an)...[rationalising]=(√a1-√a2)/(a1-a2)+
(√a2-√a3)/d...-(√an-1-√an)/d...[(tn)-(tn-1)=d=comn diff]=-(√a1-√a2)/d-
all the conecutive term will cancel excp
√an)/d=-(a1-an)/(√a1+√an)d=-(√a1-=-(a1-a1-(n-1)d)/(√a1+√an)d=(n-1)/(√a1+√an)
Alwin
1/(√a1+√a2)+1/(√a2+√a3)+.....+1/(√an-1 + √an) (√a2-√a3)/(a2-a3)+...(√an-1-√an)/({an-1}-an)...[rationalising]=(√a1-√a2)/(a1-a2)+ (√a2-√a3)/d...-(√an-1-√an)/d...[(tn)-(tn-1)=d=comn diff]=-(√a1-√a2)/d- all the conecutive term will cancel excp√an)/d=-(a1-an)/(√a1+√an)d=-(√a1-=-(a1-a1-(n-1)d)/(√a1+√an)d=(n-1)/(√a1-√an)
Kushagra Madhukar
Dear student,
Please find the solution to your problem below.
LHS =
1/(√a1+√a2) + 1/(√a2+√a3) + ..... + 1/(√an-1 + √an)
Now rationalising
=> (√a1 – √a2)/(a1 – a2) + (√a2 – √a3)/(a2 – a3) + ….....+ (√an–1 – √an)/(an–1 – an)
= – (√a1 – √a2)/d – (√a2 – √a3)/d – ...... – (√an–1 – √an)/d
[Since, ar – ar-1 = – d = – comn diff]
Now each of the consecutive term will cancel out.
Hence, the result will be = √an/d – √a1/d
Now, an – a1 = (n – 1) d
Hence, d = (an – a1)/(n – 1)
Substituting
=> (√an – √a1) x (n – 1)/(an – a1)
=> (√an – √a1) x (n – 1)/[(√an – √a1) x (√an + √a1)] {since, a2 – b2 = (a – b) (a + b)}
= (n – 1)/ (√an + √a1)
= RHS
Hence proved.
Thanks and regards,
Kushagra
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