@gnehagarg said:
5 7 16 57 244 1245 7506
7 should be 6
X1 + 1, X2 + 4, X3 + 9 and so on
@Joey_Sharma said:7 is incorrect here...the pattern followed here issecond term=(first term+1)+(1^2)third term=(second term+2)+(2^2)fourth term=(third term+3)+(3^2)fifth term=(fourth term+4)+(4^2) and so on...therefore, 7 should be replace by 5+1+1^2= 6
How 5+1+1^1=6?
@gnehagarg said:
4 2.5 3.5 6.5 15.5 41.25 126.75
6.5 should be 6.75
(X1 + 1) /2, (X2 + 2) /2, (X3 + 3) /2 and so on
i love serieshow many numbers between 1 and 1000 are such that n^2+ 3n +5 is divisible by 121??
@gnehagarg said:How 5+1+1^1=6?
again my bad..was thinking correct but wrote something else... it will be
second term=(first term*1)+1^2...
replace the + in the braces with *... and u will get correct pattern
second term=(first term*1)+1^2...
replace the + in the braces with *... and u will get correct pattern
@Gsathe89 said:how many numbers between 1 and 1000 are such that n^2+ 3n +5 is divisible by 121??
if the number is div by 121, then it would also be div by 11
n^2 - 8n + 16 + 11n - 11 should be div by 11
(n-4)^2 + 11(n-1) should be div by 11
=> n-4 should be 11k form
or n = 11k+4 form
=> (11k+4)^2 + 3(11k+4) + 5 = 121k^2+88k+16+33k+12+5 = 121k^2+121k+33 = 121k(k+1)+33
and 121k(k+1)+33 mod 121 = 33
so never div by 121
@Gsathe89 said:how many numbers between 1 and 1000 are such that n^2+ 3n +5 is divisible by 121??
n^2 + 3n + 5 = 121k
n2 + 3n + 5 - 121k = 0
D = 9 - 4(5 - 121k) = 484k - 11 = 11(44k - 1)
44k - 1 = 11^(2p+1) => no integral value of k possible
no solution
Can any one solve these questions ?
1) x^1/3+y^1/3+z^1/3 = 0 then (x+y+z)^3 = ?
2) x^(x* sqr rt (x)) = (x*sqr rt (x))^x then x = ? (x is not equal to 1,0.)
@Gsathe89 said:how many numbers between 1 and 1000 are such that n^2+ 3n +5 is divisible by 121??
Never divisible by 121
Can any one solve these questions ?1) x^1/3+y^1/3+z^1/3 = 0 then (x+y+z)^3 = ?
When x= -1, y= 1. (x+y+z)^3 = 0 ?
@missionCAT13 said:
2) x^(x* sqr rt (x)) = (x*sqr rt (x))^x then x = ?
is there any other conditions for value of x?
@albiesriram said:When x= -1, y= 1. (x+y+z)^3 = 0 ? is there any other conditions for value of x?
Yup. I have edited that post.
@Gsathe89 said:how many numbers between 1 and 1000 are such that n^2+ 3n +5 is divisible by 121??
n^2 + 3n + 5 = 121k
n^2 + 3n + (5-121k) = 0
(2n+3)^2 = 11(44k - 1)
For the above equation to give integer solutions, 44k - 1 should be expressed in 11(2x+1)
which is not possible since 44k - 1 mod 11 = 10
Hence, no solutions
@missionCAT13 said:Can any one solve these questions ?2) x^(x* sqr rt (x)) = (x*sqr rt (x))^x then x = ? (x is not equal to 1,0.)
x=9/4 ?
@missionCAT13 said:Can any one solve these questions ?
2) x^(x* sqr rt (x)) = (x*sqr rt (x))^x then x = ? (x is not equal to 1,0.)
x^(x* sqr rt (x)) = (x*sqr rt (x))^x
=> x^(x^3/2) = x^(3/2x)
=> x^3/2 = 3/2x
=> x * ( x^1/2 - 3/2 ) = 0
=> x = 9/4
@missionCAT13 said:Can any one solve these questions ?1) x^1/3+y^1/3+z^1/3 = 0 then (x+y+z)^3 = ?2) x^(x* sqr rt (x)) = (x*sqr rt (x))^x then x = ? (x is not equal to 1,0.)
1) x=y=z=0
2) x=9/4
sorry for not posting the explaintion, in office