Question 1177105
given sequence {{{2}}};{{{5}}};{{{8}}}

2.1.1 if the pattern continues, then write down the next two terms

as you can see, common difference is {{{3}}}, so next two terms are {{{11}}} and {{{14}}}


2.2.2 prove that none of the terms of this sequence are perfect squares

Let each term of the given arithmetic progression be defined as  

{{{T[n]=T[1]+(n-1)d }}}, where  {{{T[1]=2}}}  and  {{{d=T[2]-T[1]=5-2=3 }}}

{{{T[n]=2+3(n-1) }}}

{{{T[n]=3n-1 }}}

Hence, each term of the AP is of the form  {{{3n-1}}},{{{n}}} ∈ {{{N}}}. 

We know that any integer is of the form  {{{3a }}}, {{{ 3a+1}}}  or {{{ 3a-1}}}  for {{{ a}}}∈{{{Z}}}. 

Taking  {{{x=3a}}},  we have {{{x^2=9a^2=3(3a^2)=3m}}} , with  {{{m=3a^2}}}. 

Taking {{{ x=3a+1}}},  we have {{{ x^2=9a^2+6a+1=3(3a^2+2a)+1=3m+1}}},  where  {{{m=(3a^2+2a) }}}

Taking  {{{x=3a+2}}},  we have  {{{x^2=9a^2+12a+4=3(3a^2+4a+1)+1=3m+1}}},  with  {{{m=3a^2+4a+1 }}}

Hence, we can see that a perfect square is always of the form  {{{3m }}} or  {{{3m+1}}} , where  {{{m}}}  is an integer.

Hence, we can infer that a perfect square is not of the form  {{{3m-1}}}. 

Also, we have each term of the given AP to be of the form  {{{3m-1}}}. 

Hence, {{{no}}}{{{ term}}} of the given AP is a {{{perfect }}}{{{square}}}.