Let three real numbers a, b, c be in arithmet | vedclass

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(A) Since $a, b, c$ are in arithmetic progression,we have $2b = a + c$. Given the arithmetic mean of $a, b, c$ is $8$,we have $\frac{a+b+c}{3} = 8$,wh

Vedclass · Accepted Answer

$120$

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(A) Since $a, b, c$ are in arithmetic progression,we have $2b = a + c$. Given the arithmetic mean of $a, b, c$ is $8$,we have $\frac{a+b+c}{3} = 8$,which implies $a+b+c = 24$. Substituting $a+c = 2b$,we get $2b + b = 24$,so $3b = 24$,which gives $b = 8$. Then $a+c = 16$,so $c = 16 - a$. Since $a+1, b, c+3$ are in geometric progression,we have $b^2 = (a+1)(c+3)$. Substituting $b=8$ and $c=16-a$,we get $8^2 = (a+1)(16-a+3)$,which simplifies to $64 = (a+1)(19-a)$. $64 = 19a - a^2 + 19 - a$,which leads to $a^2 - 18a + 45 = 0$. Factoring the quadratic equation,we get $(a-15)(a-3) = 0$. Since $a > 10$,we must have $a = 15$. Then $c = 16 - 15 = 1$. The numbers are $a=15, b=8, c=1$. The geometric mean of $a, b, c$ is $(abc)^{1/3}$. The cube of the geometric mean is $( (abc)^{1/3} )^3 = abc = 15 	imes 8 	imes 1 = 120$.

Let three real numbers $a, b, c$ be in arithmetic progression and $a+1, b, c+3$ be in geometric progression. If $a > 10$ and the arithmetic mean of $a, b$ and $c$ is $8$,then the cube of the geometric mean of $a, b$ and $c$ is

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