Question 1025992
{{{a[n]+(a[n]+d)+(a[n]+2d)+(a[n]+3d)+(a[n]+4d)=55}}}
{{{5a[n]+10d=55}}}
{{{a[n]+2d=11}}}
{{{a[n]=11-2d}}}

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{{{a[n]^2+(a[n]+d)^2+(a[n]+2d)^2+(a[n]+3d)^2+(a[n]+4d)^2=765}}}
Substituting from above,
{{{ (11-2d)^2+((11-2d)+d)^2+((11-2d)+2d)^2+((11-2d)+3d)^2+((11-2d)+4d)^2=765}}}
{{{ (11-2d)^2+(11-d)^2+(11)^2+(11+d)^2+(11+2d)^2=765}}}
{{{ (4d^2-44d+121)+(d^2-22d+121)+(121)+(d^2+22d+121)+(4d^2+44d+121)=765}}}
{{{ 10d^2+605=765}}}
{{{10d^2=160}}}
{{{d^2=16}}}
{{{d=0 +- 4}}}
If {{{d=-4}}}, then {{{a[n]=11-2(-4)=11+8=19}}}
 which leads to {19,15,11,7,3}
If {{{d=4}}}, then {{{a[n]=11-2(4)=11-8=3}}}
 which leads to {3,7,11,15,19}