Angles of triangle in AP

$Since\quad A,B,C\quad are\quad in\quad AP,\quad \\ \\ \Rightarrow \quad 2B\quad =\quad A+C\\ \\ and\quad since\quad sum\quad of\quad angles\quad of\quad a\quad triangle\quad is\quad \pi .\\ \\ \Rightarrow \quad A+B+C=\pi \\ \\ \Rightarrow \quad 3B=\pi \quad \\ \\ \Rightarrow \quad B=\pi /3\\ \\ Now,\quad using\quad the\quad cosine\quad formula\quad :\\ \\ cosB\quad =\quad { (a }^{ 2 }+{ c }^{ 2 }-{ b }^{ 2 })/2ac\\ \\ \Rightarrow \quad 1/2\quad =\quad { (a }^{ 2 }+{ c }^{ 2 }-{ b }^{ 2 })/2ac\\ \\ \Rightarrow \quad { a }^{ 2 }+{ c }^{ 2 }-{ b }^{ 2 }=ac\\ \\ \Rightarrow \quad { a }^{ 2 }+{ c }^{ 2 }-ac={ b }^{ 2 }\\ \\ \therefore \quad Required\quad value\quad =\quad \frac { a+c }{ \sqrt { { b }^{ 2 } } } =\frac { a+c }{ b } \\ \\ Now\quad a=2RsinA,b=2RsinB,c=2RsinC\\ \\ where\quad R\quad is\quad the\quad circumradius.\\ \\ \Rightarrow \frac { a+c }{ b } =\quad \frac { 2R(sinA+sinC) }{ 2Rsin(\pi /3) } =\frac { 2sin\frac { A+C }{ 2 } cos\frac { A-C }{ 2 } }{ sin(\pi /3) } \\ \\ \\ =\frac { 2sin(\pi /3)cos\frac { A-C }{ 2 } }{ sin(\pi /3) } \quad \quad (\because \quad A+C=2B\quad and\quad B=\pi /3)\\ \\ \\ =\quad 2cos\frac { A-C }{ 2 } .\\$