{ a + b + c = 2 6 a 2 + b 2 + c 2 = 3 6 4
Let a , b , c be real numbers that follow a geometric progression (in that order) and satisfy the system of equations above.
Find b .
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Great solution! Thank you for your interest! :D
Note that a,b and c are in Geometric Progression
a = first term of GP
a r = second term of GP
a r 2 = third term of GP
a ( r 2 + r + 1 ) = 2 6
a 2 ( r 4 + r 2 + 1 ) = 3 6 4
r 4 + r 2 + 1 = ( r 2 + r + 1 ) ( r 2 − r + 1 )
a ( r 2 + r + 1 ) ∗ a ( r 2 − r + 1 ) = 3 6 4
2 6 a ( r 2 − r + 1 ) = 3 6 4
a ( r 2 − r + 1 ) = 1 4 ------>subtract this from the first equation
2 a r = 1 2
a r = 6 = b
a,b and c are in G.P.
Therefore, b=√ac→b²=ac
a²+b²+c²=364
a²+ac+c²=364
a²+2ac+c²-ac=364
(a+c)²-ac=364
By first eq, a+c is 26-b
(26-b)²-b²=364
Solving for b gives 6.
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As a , b , c is a GP we have that b = a r and c = a r 2 for some real number r = 0 . The given equations can then be written as
a ( 1 + r + r 2 ) = 2 6 , (A), and a 2 ( 1 + r 2 + r 4 ) = 3 6 4 , (B).
Squaring equation (A) yields a 2 ( 1 + 2 r + 3 r 2 + 2 r 3 + r 4 ) = 6 7 6 , and when we subtract equation (B) from this result we end up with
a 2 ( 2 r + 2 r 2 + 2 r 3 ) = 3 1 2 ⟹ a 2 r ( 1 + r + r 2 ) = 1 5 6 .
Dividing this last result by equation (A) yields b = a r = 2 6 1 5 6 = 6 .