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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

Ishan , 8 Years ago
Grade 11
anser 1 Answers
Ajay

Last Activity: 8 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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