Question 1140358
 {{{-2/3}}}......{{{-1/6}}}......{{{1/3}}}......{{{5/6}}}......{{{4/3 }}}
first find differences

........{{{3/6}}}......{{{3/6}}}......{{{3/6}}}......{{{3/6}}}

all differences same, means you have an arithmetic sequence with common difference {{{d=3/6}}}

since irst term is {{{a[1]=-2/3}}}, general formula for nth term will be

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

{{{a[n]=-2/3+(3/6)(n-1)}}}-> nth term

check:
{{{a[1]=-2/3+(3/6)(1-1)}}}
{{{a[1]=-4/6+(3/6)*0}}}
{{{a[1]=-4/6 }}}...simplify
{{{a[1]=-2/3 }}}


{{{a[2]=-2/3+(3/6)(2-1)}}}
{{{a[2]=-4/6+(3/6)}}}
{{{a[2]=-1/6 }}}

{{{a[3]=-2/3+(3/6)(3-1)}}}
{{{a[3]=-2/3+(3/6)2}}}
{{{a[3]=-2/3+1 }}}
{{{a[3]=1/3  }}}

and so on