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Let "r" = the amount added to each successive term.  

 

a1 + r = a2   and   a2 + r = a3   therefore a3 = a1 + r + r    

 

given that a1 + a3 = 5   therefore (a1) + (a1 + 2r) = 5                 (eq 1)  

 

a1 + r = a2   and   a2 + r = a3   and   a3 + r = a4   therefore a4 = a1 + r + r + r   

 

given that a2 + a4 = 6   therefore (a1 + r) + (a1 + 3r) = 6            (eq 2)   

 

subtract (eq 2) minus (eq 1)   

 

(eq 2) ...............................................  2a1 + 4r = 6    

(eq 1) ...............................................  2a1 + 2r = 5    

                                                                     2r = 1         

                                                                       r = 0.5      

 

substitute r back into (eq 2)               2a1 + (4)(0.5) = 6   

                                                                       2a1 = 4  

                                                                         a1 = 2     

 

check answer  

                          a1 = 2.0 , a2 = 2.5 , a3 = 3.0 , a4 = 3.5    

 

                          a1 +  a3 = 5                a2 +  a = 6   

                         2.0 + 3.0 = 5               2.5 + 3.5 = 6    

29.04.2024