if a, b, c are distinct positive real numbers such that `a+(1)/(b)=4,b+(1)/( c )=1,c+(1)/(d)=4` and `d+(1)/(a)=1`, then
a=c and b=d
b=d and `anec`
`ab=1 " and "cdne1`
`cd=1" and "ab=ne1`
answer: a
solution: using `a.m.gtg.m,` we have   
 `a+(1)/(b)gt2sqrt((a)/(b))b+(1)/( c )gt2sqrt((b)/( c )),c+(1)/(d)gt2sqrt(( c )/(d))" and "d+(1)/(a)gt2sqrt((d)/(a))`   
 `implies(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))gt16`   
 but,`(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))=4xx1xx4xx1=16`   
 `:." "a=(1)/(b),b=(1)/( c ),c=(1)/(d)" and "d=(1)/(a)`   
 `implies" "a=(1)/(b)=2,b=(1)/( c )=(1)/(2),c=(1)/(d)=2" and "d=(1)/(a)=(1)/(2)`   
 `implies" "a=2,b=(1)/(2),c=2" and "d=(1)/(2)`   
 `implies" "a=c" and "b=d`
a - ii, b - iv, c - i, d - iii a, b and c given;let ⇒ p(k+1) is true when p(k) is true.∴ by mathematical induction is true for all natural number n≥2. use

summation of series of natural numbers -


 

- wherein

sum of first n natural numbers



 

 and

 

summation of series of natural numbers -


 

- wherein

sum of  squares of first n natural numbers



 

 

3+7+14+24+37 15k

let tn=an2+bn+c            because difference is in a.p: [4,7,10,13]

at n=1     (i)

    n=2      (ii)

    n=3        (iii)

    (ii)-(i)        (iv)

    (iii)-(ii)       (v)

    (v)-(iv)      

                    

 

from (i) 

from (ii) 







 



      

 

put n=15   

               

               

               

               

 




option 1)

126

this option is incorrect

option 2)

122

this option is correct

option 3)

81

this option is incorrect

option 4)

119

this option is incorrect 1 a c d
2 a
3 c
4 b
5 c
i got a 100% if , passing through (1, 2), satisfies the differential equation , then let put a(x2+4)+(bx+c)(2x+1) =17put 2x+1=0a=4by equating the coefficient of x2 and constant term,a+2b=04+2b=0b=-24a+c=174×4+c=17c=1from the equation(1), we get, * If the number ends with a $0$ then there are $9$ choices for the first digit, $8$ for the second and $7$ for the third, which makes $1\times9\times8\times7= 504$ possibilities.
* If the number is even ending with something else than $0$ then there are $4$ choices for the last digit, $8$ choices for the first digit (no $0$ nor the last digit), $8$ for the second digit and $7$ for the third digit, which makes $4\times8\times8\times7 =1792$


Together, this gives $2296$ numbers with $4$ distinct digits that are even. Note that this does not allow leading $0$, as you see to want it based from the question.


Correct Answer: $B$ let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b you are correct.there is a simple mathematical reason why your answer is correct, but wyzant told me that adding it to this answer required review by wyzant.try again:p(a or b) = p(a) + p(b) - p(a and b)p(neither a nor b) = 1 - p(a or b)