sum of three numbers in continued proportion is 21 and the sum of their squares is 189 find the numbers
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Answered by
5
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
please mark it as brainliest answer please!!!
thank you!!!!!
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
please mark it as brainliest answer please!!!
thank you!!!!!
sahu22:
welcome and thank you for marking it as brainliest answer
Answered by
0
b square=a* c. equ 1
a+b+c= 21. equ 2
a^2+ b^2+c^2=189. equ 3
put eq 1 in eq 3
a^2+ a*c+c^2=189
a^2+2*a*c+c^2=189+a*c
(a+c)^2= 189+b^2
(21-b)^2= 189+b^2
441 - 42b+ b^2= 189+b^2
b=6
.....
get other values by putting in eq above
a+b+c= 21. equ 2
a^2+ b^2+c^2=189. equ 3
put eq 1 in eq 3
a^2+ a*c+c^2=189
a^2+2*a*c+c^2=189+a*c
(a+c)^2= 189+b^2
(21-b)^2= 189+b^2
441 - 42b+ b^2= 189+b^2
b=6
.....
get other values by putting in eq above
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