Question

help please I tried 4 times
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Answer 1

Answer:

a, b, c are in continued proportion

=> a/b = b/c => b² = ac

So

( a + b + c ) ( a - b + c )

= ( a + c )² - b²

= a² + 2ac + c² - ac

= a² + ac + c²

= a² + b² + c²

=> ( a² + b² + c² ) / ( a + b + c ) = a - b + c

Now a² + b² + c² = 133 and a + b + c = 19

=> a - b + c = 133 / 19 = 7

Thus

2b = ( a + b + c ) - ( a - b + c ) = 19 - 7 = 12  =>  b = 6

This leaves   a + c = 19 - 6 = 13   and   a² + c² = 133 - 6² = 97.

Taking a = 4 and c = 9 finishes this off.

So a = 4, b = 6, c = 9 have sum 19 and sum of squares 133.

Answer 2

mujhe nahi aa Raha h