Question

2+4+6...+n/1+3+5...+n=37/46
find n
​

Answer 1

Answer:

doubtnut is very helpful app

Step-by-step explanation:

download the app

Answer 2

Step-by-step explanation:

Sn=n/2[2a+(n-1)d]

Here,

Sn= 37/46

a=2

d=4-2=2

So,

[n/(n/2(2+(n-1)2)]/2[4+n/(n/2(2+(n-1)2))-1)2=37/46

=>[n/n^2/2{4+n^2-1}2=37/46

=>1/n[3+n^2]2=37/46

=>(6+2n^2)/n=37/46

=>46(6+2n^2)=37×n

=>276+92n^2=37n

=>92n^2-37n+276=0

Solving this quadratic equation by quadratic formula,we have

a=92, b=-37, c=276

So,

-b+-√b^2-4ac/2a

37+-√37^2-4×92×276/2×92

Hence, the ans will be found by solving this.....in root