Question 1178676


Find the nth partial sum of the arithmetic sequence.

given:

{{{a[n] = 2n -x}}}
 {{{n = 19}}}


the nth partial sum formula of the arithmetic sequence:

{{{S[n]=(n/2)*(2a[1]+(n-1)d)}}}


find first term:

{{{a[n]=a[1]+d(n-1)}}}

so,  {{{a[1]+d(n-1)= 2n -x}}}

substitute {{{n = 19}}}

=>{{{a[1]+d(19-1) = 2*19 -x}}}

=>{{{a[1]+18d= 38 -x}}}

=>{{{a[1]=  38 -x-18d}}}


go to


{{{S[n]=(n/2)*(2a[1]+(n-1)d)}}}.........substitute {{{a[1] }}}and {{{n}}}

{{{S[19]=(19/2)*(2(38 -x-18d)+(19-1)d)}}}

{{{S[19]=(19/2)*(76 -2x-36d+18d)}}}

{{{S[19]=(19/2)*(-18 d - 2 x + 76)}}}

{{{S[19]=-171d - 19x + 722}}}