Question 1178050

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


if 6th term {{{S[6]=35}}} and  13term is {{{S[13]=77}}}, we have


{{{35=S[1]+(6-1)*d}}}
{{{35=S[1]+5d}}}.......solve for {{{S[1]}}}
{{{S[1]=35-5d}}}.............eq.1


{{{77=S[1]+(13-1)*d}}}
{{{77=S[1]+12d}}}.......solve for {{{S[1]}}}
{{{S[1]=77-12d}}}.............eq.2


equate right sides of eq.1 and eq.2


{{{35-5d=77-12d}}}.....solve for {{{d}}}
{{{12d-5d=77-35}}}
{{{7d=42}}}
{{{d=6}}}


go to

{{{S[1]=35-5d}}}.............eq.1, substitute {{{d}}}
{{{S[1]=35-5*6}}}
{{{S[1]=5}}}


so, your n-th term formula is:


{{{S[n]=5+(n-1)*6}}}


now calculate  {{{S[4]}}} 

{{{n=4}}}
{{{S[4]=5+(4-1)*6}}}
{{{S[4]=5+3*6}}}
{{{S[4]=23}}}


since given 6th term {{{S[6]=35}}}


then  


{{{S[6]-S[4]=35-23=12}}}