Math, asked by ADITYABHAKAT12345, 1 year ago

sum of three numbers in continued proportion is 21 and the sum of their squares is 189 find the numbers

Answers

Answered by sahu22
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 

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sahu22: welcome and thank you for marking it as brainliest answer
Answered by yuvrajadkarp8eygz
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
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