<PRE><B>Question 878931</B>
the sum of three of a g.p. is 21 and the sum of their squares is 189. find the numbers.
let three terms are   a,b,c in GP
as per given conditions 
a+b+c =21(1)
a+c=21-b
a^2 + b^2 + C^2 = 189(2)
by using formula 
(a+b+C)^2= a^2 + b^2 +c^2 + 2ab +2bc+2ca by substituting the value         from (1) and (2)
(21)^2= 189+2(ab +bc +ca )
441 =189 +2(ab +bc +ca )
441-189 /2= ab + bc +ca 
252/2 =ab+bc +ca 
126= ab +bc +ca   we know if a,b,c are in GP than b^2=ac
126=ab +bc +b^2
126=b(a+c) +b^2 by putting a+c=21-b from (1)
126=b(21-b) +b^2
126= 21b -b^2 +b^2
126=21b
126/21 =b
b=6
a+c=21-b
a+c=21-6 =15 
both a and c are in  GP than a=3 and c=12
therefore 
Answer a=3    b=6   c= 12 







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