Question 723103
Let the number in arithmetic sequence with a common difference of 4 be
{{{x-4}}} , {{{x}}}, and {{{x+4}}}
I assume that by "the first number" the problem means the smallest of them, which would be {{{x-4}}}
If {{{x-4}}} is increased by 2, we get
{{{x-4+2=x-2}}}.
If {{{x}}} (the second number) is increased by 3 , we get
{{{x+3}}}.
If {{{x+4}}} (the third number) is increased by 5, we get
{{{x+4+5=x+9}}}
Since the numbers {{{x-2}}}, {{{x+3}}}, and {{{x+9}}} form a geometric sequence, the ratio of one number to the next is the same, meaning that
{{{(x+9)/(x+3)=(x+3)/(x-2)}}}
Equating the cross products (or, if you prefer, multiplying both sides of the equal sign times {{{(x-2)(x+3)}}} to eliminate denominators) we get
{{{(x+9)(x-2)=(x+3)^2}}} --> {{{x^2+7x-18=x^2+6x+9}}} --> {{{7x-18=6x+9}}} --> {{{7x-6x=9+18}}} --> {{{highlight(x=27)}}}
The original numbers are:
{{{x-4=27-4=highlight(23)}}}
{{{x=highlight(27)}}} and
{{{x+4=27+4=highlight(31)}}}