Question

find three numbers are in g.p such that their sum is 21and sum of their square is 189.​

Answer 1

let three terms are a,b,c in GP

a+b+c =21(1)

a+c=21-b

a^2 + b^2 + C^2 = 189(2)

(a+b+C)^2= a^2 + b^2 +c^2 + 2ab +2bc+2ca

(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

b=6

a+c=21-b

a+c=21-6 =15 

So, a=3 b=6 c= 12

Answer 2

Step-by-step explanation:

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