1x3+x4= r^2*(x1+x2) so r=±2 so x1(1+r)=3 implies, x1=1 or -3 then x2=2 or 6 then x3 and x4 are (4 or -12) and (8 or 24) so p=2 or -18 q=32 or -288
let x1,x2 be the roots of the eq x2-3x+p and x3,x4 be the roots of the eq x2-12x+q =0,then no's x1,x2,x3,x4 are in G.P then value of p,q
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3 Answers
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1u c, x1=a, x2=ar, x3=ar^2, x4=ar^3.
so from first eqn, a(1+r)=3 --------------1,
a^2r=p ------------------2
from second, ar^2(1+r) = 12---------------3,
a^2r^5=q --------------4
divide 3 by 1...
so u have r^2=4 =>r = +-2
now wen u got r, u will get a..
then u will get p and q...