Question 955773
given:
{{{P=30}}} units
The length of the sides form an arithmetic sequence.


 let {{{a}}} be the shortest side therefore the first term
let {{{r}}} be the common difference

the series is therefore:
{{{a}}}, {{{a+r}}}, {{{a+2r}}}

then {{{P=a+a+r+a+2r}}}

{{{30=3a+3r}}}

{{{30=3(a+r)}}}

{{{30/3=a+r}}}

{{{10=a+r}}}


all possible sets of the lengths of the sides of the triangle:


if {{{a=9}}} and {{{r=1}}}, then the lengths of the sides of the triangle are

{{{a=9}}}
{{{a+r=a+1=10}}}
{{{a+2r=a+2=11}}}.........first solution: possible sides are  {{{highlight(9)}}},{{{highlight(10)}}},and {{{highlight(11)}}}

if {{{a=8}}} and {{{r=2}}}, then the lengths of the sides of the triangle are

{{{a=8}}}
{{{a+r=8+2=10}}}
{{{a+2r=8+4=12}}} .........2nd solution: possible sides are  {{{highlight(8)}}},{{{highlight(10)}}},and {{{highlight(12)}}}

if {{{a=7}}} and {{{r=3}}}, then the lengths of the sides of the triangle are

{{{a=7}}}
{{{a+r=7+3=10}}}
{{{a+2r=7+6=13}}}.........3rd solution: possible sides are  {{{highlight(7)}}},{{{highlight(10)}}},and {{{highlight(13)}}}

if {{{a=6}}} and {{{r=4}}}, then the lengths of the sides of the triangle are

{{{a=6}}}
{{{a+r=6+4=10}}}
{{{a+2r=6+8=14}}}.........4th solution: possible sides are  {{{highlight(6)}}},{{{highlight(10)}}},and {{{highlight(14)}}}

if {{{a=5}}} and {{{r=5}}}, then the lengths of the sides of the triangle are

{{{a=5}}}
{{{a+r=5+5=10}}}
{{{a+2r=5+10=15}}}.........5th solution: possible sides are  {{{highlight(5)}}},{{{highlight(10)}}},and {{{highlight(15)}}}

if {{{a=4}}} and {{{r=6}}}, then the lengths of the sides of the triangle are

{{{a=4}}}
{{{a+r=4+6=10}}}
{{{a+2r=4+12=16}}}.........6th solution: possible sides are  {{{highlight(4)}}},{{{highlight(10)}}},and {{{highlight(16)}}}

if {{{a=3}}} and {{{r=7}}}, then the lengths of the sides of the triangle are

{{{a=3}}}
{{{a+r=3+7=10}}}
{{{a+2r=3+14=17}}}.........7th solution: possible sides are  {{{highlight(3)}}},{{{highlight(10)}}},and {{{highlight(17)}}}

if {{{a=2}}} and {{{r=8}}}, then the lengths of the sides of the triangle are

{{{a=2}}}
{{{a+r=2+8=10}}}
{{{a+2r=2+16=18}}}.........8th solution: possible sides are  {{{highlight(2)}}},{{{highlight(10)}}},and {{{highlight(18)}}}

if {{{a=1}}} and {{{r=9}}}, then the lengths of the sides of the triangle are

{{{a=1}}}
{{{a+r=1+9=10}}}
{{{a+2r=1+18=19}}} .........9th solution: possible sides are  {{{highlight(1)}}},{{{highlight(10)}}},and {{{highlight(19)}}}