Question 1199566
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Let x be the common difference in the arithmetic sequence.  Then....<br>
2, b, and c form an arithmetic sequence, so
{{{b = 2+x}}}
{{{c = 2+b = 2+2x}}}<br>
b, c, and d form a geometric sequence, so
{{{bd = c^2}}}
{{{d = c^2/b}}}<br>
c, d, and e form a harmonic sequence, so
{{{2/d=1/c+1/e}}}
{{{2/d-1/c=1/e}}}
{{{(2c-d)/cd=1/e}}}
{{{e=cd/(2c-d)}}}<br>
So we have
{{{b=x+2}}}
{{{c=2x+2}}}
{{{d=c^2/b=(2x+2)^2/(x+2)}}}
{{{e=cd/(2c-d)=((2x+2)^3/(x+2))/(2(2x+2)-((2x+2)^2)/(x+2))}}}<br>
We need to have e = 72.  Simplify the expression for e, starting by multiplying numerator and denominator of the fraction by (x+2).<br>
{{{e=((2x+2)^3)/((2(2x+2)(x+2)-(2x+2)^2))}}}<br>
{{{e=((2x+2)^3)/((2x+2)(2(x+2)-(2x+2)))}}}<br>
{{{e=(2x+2)^3/((2x+2)(2))}}}<br>
{{{e=(2x+2)^2/2 = 4(x+1)^2/2 = 2(x+1)^2}}}<br>
Then solve for e=72.<br>
{{{2(x+1)^2=72}}}
{{{(x+1)^2=36}}}
{{{x+1=6}}}
{{{x=5}}}<br>
And we have for the sequence of numbers:
2
b = 2+5 = 7
c = 7+5 = 12
d = 12^2/7 = 144/7
e = (12(144/7))/(24-144/7) = (1728/7)/(24/7) = 72<br>
ANSWER: b+c = 7+12 = 19<br>