Question 726305


If {{{x}}} is any {{{odd}}} number, then {{{x}}} and {{{x + 2}}} are consecutive odd numbers.

you need four consecutive odd numbers, and they are:{{{x}}},{{{x + 2}}}, {{{x + 4}}}, and {{{x + 6}}}

given:

{{{3}}} consecutive odd numbers whose sum is {{{5}}} more than twice the next (fourth) consecutive odd number

so, you have

{{{x+(x + 2)+(x + 4)=2(x + 6)+5}}}

{{{x+x + 2+x + 4=2x + 12+5}}}

{{{3x + 6=2x + 17}}}

{{{3x-2x=17-6}}}

{{{x=11}}}......first odd number

{{{x+2=13}}}.....second odd number

{{{x+4=15}}}.....third odd number

and

{{{x+6=17}}}.....fourth odd number