Question 1181280


The first four terms of an arithmetic sequence are 

{{{2}}}, {{{a-b}}}, {{{2a+b+7}}}, {{{a-3b }}}

=> first term is{{{ 2}}}

in an arithmetic sequence we add common difference {{{d}}} to get next term, so

{{{2+d=a-b}}}....eq.1

add common difference {{{d}}} to second term to get third term
{{{a-b+d= 2a+b+7}}}.......eq.2

add common difference {{{d}}} to  third term to get fourth term

{{{2a+b+7+d= a-3b}}} ..........eq.3


solve the system

{{{2+d=a-b}}}....eq.1
{{{a-b+d= 2a+b+7}}}.......eq.2
{{{2a+b+7+d= a-3b}}} ..........eq.3
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{{{2+d=a-b}}}....eq.1, solve for {{{d}}}
{{{d=a-b-2}}}........eq.1a

{{{a-b+d= 2a+b+7}}}.......eq.2, solve for{{{ d}}}
{{{d= 2a+b+7-a+b}}}
{{{d= a+2b+7}}} ................eq.2a


from eq.1a and eq.2a we have

{{{a-b-2=a+2b+7}}}............., solve for {{{b}}}
{{{a-7-2-a=2b+b}}}
{{{-9=3b}}}
{{{b=-3}}}

go to

{{{d=a-b-2}}}........eq.1a, substitute{{{ b}}}
{{{d=a-(-3)-2}}}
{{{d=a+3-2}}}
{{{d=a+1}}}......................eq.1b


go to

{{{2a+b+7+d= a-3b}}} ..........eq.3, substitute {{{b}}} and {{{d}}}

{{{2a-3+7+a+1= a-3(-3)}}}
{{{3a+5= a+9}}}
{{{3a-a= 9-5}}}
{{{2a= 4}}}
{{{a=2}}}


go to

{{{d=a+1}}}......................eq.1b, substitute {{{a}}}
{{{d=2+1 }}}
{{{d=3 }}}


answer: {{{a=2}}}, {{{b=-3}}}, common difference {{{d=3}}}


and your terms are:
{{{2}}}, {{{2-(-3)}}}, {{{2*2-3+7}}}, {{{2-3(-3) }}}

{{{2}}}, {{{5}}}, {{{8}}}, {{{11 }}}