IIT JEE , AIEEE Forum - Mathematics , Algebra

rishipratimm

suppose there are integers from 1 to 2n+1.three numbers are selected at random.then find the probability that they are in a.p.

rishipratimm

hey guys...
can any of u help????????

PORSHE

when the common differce is r no of favourable cases=2n\r-1

required probability =summation (2n\r-1)(r=1to r=n)\2n+1c3

abhijeet_0201

let a , b ,c be the nos

then a , c shud be either both odd or both even (since 2b=a+c and 2b is always even ans sum of 2 nos is even only when they r both odd or even)
so no of fav cases=n+1C2 + nC2
prob=(n+1C2+nC2)/2n+1C2

=n/(2n+1)

is that the ans????????

saurabh_reincarnated

but i am getting diff. ans............

how abt. doing this way......

let middle term=a

other terms r... a+d and a-d

for a= 1, d can assume 0 val.

a=2, d can be 1, i value

....... 4 a=2n+1.,d=0.................. no.of values of d will be repeated....

add thm all and divide it by total no. of cases....... we get prob.

addin wont be a prob. as they r in series........

i am getting ans.= 3n/(4n^2-1)

is it rit?

hope this wrks

Cheers!!!!!!!!

Greatdreams

Let the integers be 1,2,3......2n-1

so we have then no of ways to select 3 numbers = ²ⁿ⁺¹C₃

So let us find the possible A.Ps

1.)common diff = 1

(1,2,3),(2,3,4)......(2n-1,2n,2n+1)

So no. of ways = 2n-1

2.)common diff = 2

(1,3,5).........(2n-3,2n-1,2n+1)

so no of ways = 2n - 3

So possible A.P with common diff = n is , (1,n+1,2n+1)

So n(E) = (2n-1)+(2n-3)+(2n-5)...+1

= 1+3+5+...+(2n-1)

=n/2{2*1+(n-1)*2} = n²

So probability = n(E)/n(S) = n²/²ⁿ⁺¹C₃ = 3n/4n²-1

saurabh_reincarnated

well i was also getti that ans.......

its rit then........

and gr8 dreams the methud u and i hve applied seems to be almost of 1 kind....

but what abhijeet did seems a gud methud........but there are flaws .......is there some way we can modify that.........i thnk this approach can be used in solvin quite a dew problems.....

isnt it...........so do send anoder methud!111

cheers!!!!!!!!!

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