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@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Looks like some componendo dividendo stuff. Yum
2q/p?
Well, assume the terms to be a,ar..., then a+ar=3 and ar^2+ar^3=6; [sum of roots of quadratic] which means common ratio is sqrt(2) and a=3/(sqrt(2)+1)
Using this, you can find the value of p and q, and hence, find the ratio.
how is r= 2 , wont it be root 2
r(a+ar)/(a+ar)=6/3=2.
alpha (1+r) = 3
alpha r^2 (1+r ) 6
dividing those r^2 = 2
Oh right, my bad.
also i cant get answer , that ratio is componenedo dividendo so it will simpify to 2q/p right?
If the ratio is lambda, dont we do componendo dividendo both sides?
ooh thats why i wasnt getting got it
ye ans should be 3
AHHHH THIS IS WHAT I WAS MISSING
OK
I WAS MISSING 2 EQATIONS GOT THEM
THANKS
I WONT EVEN READ THE MESSAGES BELOW
+solved @SirLancelotDuLac
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