Find two consecutive natural numbers , the sum of whose reciprocal is 11 / 30. Please do solve it with full explanation. FAST !!!
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let x, x+1 are two consecutive natural numbers
sum of their reciprocals =11/30
1/x+1/(x+1)=11/30
(x+1+x)/[x(x+1)] =11/30
(2x+1)/(x^2+x)=11/30
30(2x+1)=11(x^2+x)
60x+30=11x^2+11x
11x^2+11x-60x-30=0
11x^2-49x-30=0
11x^2-55x+6x-30=0
11x(x-5)+6(x-5)=9
(x-5)(11x+6)=0
x-5=0 or 11x+6=0
x=5 or x=-6/11
but x is a natural number
therefore x=5
the required numbers are 5,6
sum of their reciprocals =11/30
1/x+1/(x+1)=11/30
(x+1+x)/[x(x+1)] =11/30
(2x+1)/(x^2+x)=11/30
30(2x+1)=11(x^2+x)
60x+30=11x^2+11x
11x^2+11x-60x-30=0
11x^2-49x-30=0
11x^2-55x+6x-30=0
11x(x-5)+6(x-5)=9
(x-5)(11x+6)=0
x-5=0 or 11x+6=0
x=5 or x=-6/11
but x is a natural number
therefore x=5
the required numbers are 5,6
Anonymous:
Thank you mysticd for your answer.
ur welcome
Can I mark it as brainliest answer ?
yes
But I can't see the brainliest answer option there.
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