question number 14 please - math - quadratic equations
question number 14 please
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please 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 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` fill in the blank (xiv). given (pq – prq) = (pq×1 – pq×r)taking pq common,(pq – prq) = pq(1 – r) solution: `1 + omega + omega^2 = 0`   

`z, z omega , z + z omega`   

`z , z omega , z(1 + omega)`   

`z , z omega , - z omega^2 `   

area `/\_ abc= 1/2 xx b xxh `   

`1/2 xx ab xx od`   

`1/2 xx 1 xx sqrt3/2 = sqrt3/4`   

so `1, omega , -omega^2`   

`z, z omega , - z omega^2 `   

area = `|z|^2 xx sqrt3/4 = 4/sqrt3`   

`|z|^2 = 16/3`   

`|z| = 4/ sqrt3`   

answer answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a 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) the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is there are 2 pints in a quart. there are 4 quarts in a gallon. 1 quart = 2 pints