Algebra

 

 

Let a1, a2, a3, ..., a10 be an arithmetic series. If a1 + a3 = -2 and a2 + a4 + a6 = 1, then find a1.    

 

call "d" the amount of addition between successive terms.    

 

starting with a₁   then  a₂  =  a₁ + d                                              =   a₁ + d     

                                    a₃  =  a₂ + d  =  (a₁ + d) + d                     =   a₁ + 2d    

                                    a₄  =  a₃ + d  =  (a₁ + d + d) + d               =   a₁ + 3d    

                                    a₅  =  a₄ + d  =  (a₁ + d + d + d) + d         =   a₁ + 4d    

                                    a₆  =  a₅ + d  =  (a₁ + d + d + d + d) + d   =   a₁ + 5d    

 

given       a₁ + a₃  =  – 2    

                a₁ + a₁ + 2d  =  – 2    

                2a₁ + 2d  =  – 2    

                     a₁ + d  =  – 1                                   (eq 1)    

 

given        a₂ + a₄ + a₆  =  1    

                (a₁ + d) + (a₁ + 3d) + (a₁ + 5d)  =  1    

                3a₁ + 9d  =  1                                       (eq 2)    

 

multiply (eq 1) by – 9         – 9a₁ – 9d  =  9    

add (eq 2)                             3a₁ + 9d  =  1    

                                           – 6a₁          =  10    

                                                a₁          =  – 10 / 6    

                                                        a₁  =  – 5 / 3    

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