Question 1203707


The formula for a sequence of consecutive integers is {{{n}}}, {{{n+1}}}, {{{n+2}}},{{{ n+3}}}, {{{n+4}}},..., {{{n+k}}}.

The sum of the first and {{{4 }}}times the third is equal to {{{56}}} less than {{{3 }}}times the sum of the second, fourth, and fifth.

{{{n+4(n+2)=3((n+1)+(n+3)+(n+4))-56}}}

{{{n+4n+8=3(3n+8)-56}}}

{{{5n+8=9n+24-56}}}

{{{56+8-24=9n-5n}}}

{{{40=4n}}}

{{{n=10}}}

 five consecutive integers are:{{{10}}}, {{{11}}}, {{{12}}}, {{{14}}}, {{{14}}}

check:

{{{10+4(12)=3((11)+(13)+(14))-56}}}
{{{10+48=3(38)-56}}}
{{{58=58}}}