Find the respective terms for the following APs a1-4-a6-612 find a2-a3-a4-a5

a_1= 4 #160; #160; a_6=6 a_1=a= 4 #160; ...........(1) a_4=a+5d=6 .......(2)Equation (1) (2), gives 5d= 10 ⇒ d=2 ∴ a_ ...

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Find the respective terms for the following $A P s$
$a_{1} = - 4, a_{6} = 6 \frac{1}{2}$ find $a_{2}, a_{3} . a_{4}, a_{5}$

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A P s

a_{1} = 5, a_{4} = 9 \frac{1}{2}

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a_{1} = 2; a_{3} = 26

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a_{1}, a_{2}, a_{3},

\frac{a_{4}}{a_{7}} = \frac{2}{3}

\frac{a_{6}}{a_{8}}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}

(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =

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