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Q6 of the book please Q6 of the book please Q6 of the book please Q6 of the book please tell

Q6 of the book please Q6 of the book please Q6 of the book please Q6 of the book please tell

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Grade:11

1 Answers

Ajay
209 Points
7 years ago
Solution below, but do let us know where you found difficulty..............................................
Given\quad a\quad b\quad c\quad are\quad in\quad HP\quad \\ Implies\quad 2ac\quad =\quad b(a+c)\quad -----------------(1)\\ log\quad (a+c)\quad +\quad log(a-2b+c)\\ \quad =\quad log(a+c)(a+c-2b)\\ put\quad b\quad =\frac { 2ac }{ (a+c) } \quad from\quad (1)\\ =\quad log\left[ (a+c)\left\{ (a+c)-\frac { 4ac }{ (a+c) } \right\} \right] \quad \\ =log\left[ { (a+c) }^{ 2 }\quad -4ac\quad \right] \\ =log{ (a-c) }^{ 2 }\\ since\quad c>a\\ =\quad log{ (c-a) }^{ 2 }\\ =2log(c-a)\\

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