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b1+b2+b3=62

b1^2+b2^2+b3^2=2604

b1+b1q+b1*q^2=62

b1^2+b1^2*q^2 +b1^2*q^4=2604

 x+xy+xy^2=62

x^2+(xy)^2+x^2*y^4=2604

 x(1+y+y^2)=62

 x^2(1+y^2+y^4)=2604

 первое на второе поделим

1+y +y^2/ 1+y^2+y^4=x/42

 42(1+y+y^2)=x(1+y^2+y^4)

42(1+y+y^2)=x(1+y^2+y^2^2)

42/x= y^2-y+1

{xy^2-yx+x=42

{  x+xy+xy^2=62

{xy^2=42 +yx-x

{ xy^2=62-xy-x

{42+yx-x=62-xy-x

{2yx=20

{yx=10

{y=10/x

{x+xy+xy^2=62    ставим

{ x+10+100/x=62

{x^2+10x-62x+100=0

{ x^2-52x   +100 =0

 x=2

 x2=50

значит b1=2  and b1=50 

q=5

q=1/5 убывающая

 значит b1=2

 b2=10

b3=50 

Проверим 50^2+10^2+2^2  =2604

ответ   

b1=2

b2=10

b3=50