Question 1135842


Let the three terms of AP are {{{a}}},{{{a+d}}},{{{a+2d}}}.
Now,

As per the question ,
 
{{{a+a+d+a+2d =21}}}

{{{ 3a+3d = 21 }}}

{{{ a+d = 7 }}}

=>{{{ a+d = 7}}} => {{{d= 7-a}}}....eq.1

Now, 

{{{a(a+2d) =(a+d)+ 6}}}

{{{a(a+2d)-(a+d) = 6}}}

{{{a^2+2ad-a-d = 6}}}.........substitute {{{d}}} from eq.1

{{{a^2+2a(7-a)-a-(7-a) = 6}}}

{{{a^2+14a-2a^2-a-7+a = 6}}}

{{{-a^2+14a = 6+7}}}

{{{ a^2-14a +13=0}}}...factor

{{{ a^2-a-13a +13=0}}}

{{{ (a^2-a)-(13a -13)=0}}}

{{{ a(a-1)-13(a -1)=0}}}

{{{ (a-13)(a -1)=0}}}

so you will get

{{{a= 13}}}, or {{{a=1}}}


find {{{d}}}


{{{d= 7-a}}}....eq.1

{{{d= 7-13}}}

{{{d= -6}}}

or

{{{d= 7-a}}}....eq.1

{{{d= 7-1}}}

{{{d= 6}}}



Now, there will be two AP:

{{{13}}},{{{7}}}, {{{1}}} and {{{1}}},{{{7}}},{{{13}}}