Question 823499
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Hi,
The sum of the first n terms of the Arithmetic Progression  13, 16.5 , 20  
is the same as the sum of the first n terms of the Arithmetic progression 3, 7, 11   
calculate n.
{{{S[n]=(n/2)(2*a[1]+ (n-1)d)}}}
{{{(n/2)(26 + (n-1)3.5) =(n/2)(3 + (n-1)4)  )}}} |Multiplying thru by 2/n
{{{26 + (n-1)3.5 = 3 + (n-1)4  }}}
       26 + 3.5n - 3.5 = 3 + 4n - 4
         26  - 3.5 +1= .5n
                 23.5 = .5n 
                     n = 47  
   CHECKING our answer***
 {{{26 + (n-1)3.5 = 3 + (n-1)4  )}}}
 {{{ 26 + (46)3.5 = 3 + (46)4  }}}
      {{{ 187 = 187)}}}