qi)

then   Ans =

q2)

  and 

then     Ans =

q3)

   Ans  =

Q3 is very easy,As n tends to infinity it is the ratio of two infinite sums,So,

1/(1-1/2):1/(1-1/3)2:3/2=4/3.Wait for the other solns.

Q2.S_n=(-1)^r(2/3)[(1/r)+(1/r+1)]=(2/3)[(-1+1/2-1/3+1/4-1/5+.....)+(-1/2+1/3-1/4+1/5-1/6+....)].Now as n tends to infinity,the terms cancel and sigma tends to -1.Hence the ans is -2/3

here is the soln of 1st ques.

solve the determinant as follows.

i dont know how to write the determinant here thats y i m explaining the whole method.

expand the sin(a+b)=sinacosb+cosasinb in all the terms having this form

then put cosxsinyh=0 as h tends to 0

u will get sina, sinacos2h, and sinacosh in 1 column and sinacos2h,sinacosa and sina in 3 column

C₁C₃-C₁

 then take out sina common from the i column

we will get 2sin²h,2sin(h/2)sin(3h/2) and sin²(h/2)

now as h0   sinx=x

take out 2 common from 1 column and the put sinh=h

and take out h²   common from 1 column and then solve the determinant we will get 

lim  (h)/h²=9sincos²

hope u got it........

@allamraju .........can u explain the third ques wid the eqns...plz becoz i am gettin confused wid that ratio and fractions....

The third question is very simple.Because I don't know how to introduce fraction symbols,I denoted like that.I think U know the formula of sum of infinite G.P which is

S=a/1-r,In the nr.,a=1 and r=1/2 and in the dr.,a=1 and r=1/3.So,nr. is 1/1-1/2=2 and dr. is 1/1-1/3=3/2.Hence the ans is 4/3