JEE Mains · Maths · STD 11 - 8. sequence and series
For any three positive real numbers \(a,b,c\) ; \(9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)\) then
- A \(a,b,c\) are in \(G.P.\)
- B \(b,c,a\) are in \(G.P.\)
- C \(b,c,a\) are in \(A.P.\)
- D \(a,b,c\) are in \(A.P.\)
Answer & Solution
Correct Answer
(C) \(b,c,a\) are in \(A.P.\)
Step-by-step Solution
Detailed explanation
We have \(9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)\) \( \Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc\) \( \Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + 5{c^2} - 75ac - 45ab - 15bc = 0\)…
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