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Joint Entrance Examination
Mathematics
Sequences and Series
Question 434
Question 434 of 519
Medium
MCQ
For any three positive real numbers a, b and c, 9(25
a
2
{a^2}
a
2
+ b2) + 25(c2 - 3
a
a
a
c) = 15b(3
a
a
a
+ c). Then
A
b, c and \[a\] are in G.P.
B
b, c and \[a\] are in A.P.
C
\[a\], b and c are in A.P.
D
\[a\], b and c are in G.P.
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Official solution
9(25
a
2
{a^2}
a
2
+ b2) + 25(c2 - 3
a
a
a
c) = 15b(3
a
a
a
+ c) \[ \Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc\] \[ \Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + {\left( {5c} \right)^2} - 75ac = 45ab + 15bc\] \[ \Rightarrow \] \[{1 \over 2}\left[ {{{\left( {15a - 3b} \right)}^2} + {{\left( {3b - 5c} \right)}^2} + {{\left( {5c - 15a} \right)}^2}} \right] = 0\] it is possible when 15a -- 3b = 0, 3b -- 5 c = 0 and 5c -- 15a = 0 \[ \Rightarrow \] 15a = 3b = 5c \[ \Rightarrow \] b = \[{{5c} \over 3}\], a = \[{c \over 3}\] \[ \Rightarrow \] a + b = \[{c \over 3} + {{5c} \over 3}\] = \[{{6c} \over 3}\] = 2c \[ \therefore \] b, c, a are in A.P.
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