Question 1131474


 an arithmetic progression 

{{{2}}}, {{{-1}}},{{{-4}}},.... 

first term: {{{a[1]=2}}}
{{{2-3=-1}}}
{{{-1-3=-4}}}

=>common difference:{{{d=-3}}} 

nth term:
{{{a[n]=a[1] +(n-1)d}}}
{{{a[n]=2 -3(n-1)}}}

then,
{{{a[4]=2 -3(4-1)}}}
{{{a[4]=2 -3(3)}}}
{{{a[4]=2 -9}}}
{{{a[4]= -7}}}

{{{a[5]=2 -3(5-1)}}}
{{{a[5]=2 -3(4)}}}
{{{a[5]=2 -12}}}
{{{a[5]= -10}}}
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your sequence:

{{{2}}}, {{{-1}}}, {{{-4}}}, {{{-7}}}, {{{-10}}}, {{{-13}}}, {{{-16}}}, {{{-19}}}, {{{-22}}}, {{{highlight(-25)}}}, {{{highlight(-28)}}}, {{{highlight(-31)}}}, {{{-34}}}, {{{-37}}}, {{{-40}}}, {{{-43}}}, {{{-46}}}, ...

state three consecutive terms in this progression which sum up to {{{-84}}}:

The three consecutive terms are {{{a}}}, {{{a - 3}}}, and {{{a -6}}}.


{{{a + (a - 3) + (a - 6) = -84}}}

{{{a = -25}}}

then three consecutive terms are:

{{{-25}}}, {{{-28}}}, {{{-31}}}

check:
{{{(-25)+(-28)+(-31)=-84}}}

{{{-84=-84}}}