Consider two positive numbers a and b. If arithmetic mean of a and b exceeds their geometric mean by (3/2) and geometric mean of a and b exceeds their harmonic mean by (6/5), then the absolute value of (a2-b2) is equal to

Question

Consider two positive numbers a and b. If arithmetic mean of a and b exceeds their geometric mean by (3/2) and geometric mean of a and b exceeds their harmonic mean by (6/5), then the absolute value of (a2-b2) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A=G+(3/2), G=H+(6/5) Using G2=A H we get G=6, A=(15/2) and H=(24/5) Hence, a+b=15 and a b=36 a=12, b=3 or a=3, b=12 |a2-b2|=135.

Q. Consider two positive numbers $a$ and $b$ . If arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$ , then the absolute value of $(a^{2} - b^{2})$ is equal to

$A = G + \frac{3}{2}, G = H + \frac{6}{5}$
Using $G^{2} = A H$ we get
$G = 6, A = \frac{15}{2}$ and $H = \frac{24}{5}$
Hence, $a + b = 15$ and $ab = 36$
$a = 12, b = 3$ or $a = 3, b = 12$
$∣ ∣ a^{2} - b^{2} ∣ ∣ = 135$ .

Q. Consider two positive numbers a and b. If arithmetic mean of a and b exceeds their geometric mean by 23​ and geometric mean of a and b exceeds their harmonic mean by 56​, then the absolute value of (a2−b2) is equal to

A=G+23​,G=H+56​ Using G2=AH we get G=6,A=215​ and H=524​ Hence, a+b=15 and ab=36 a=12,b=3 or a=3,b=12 ∣∣​a2−b2∣∣​=135.

Q. Consider two positive numbers $a$ and $b$ . If arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$ , then the absolute value of $(a^{2} - b^{2})$ is equal to

$A = G + \frac{3}{2}, G = H + \frac{6}{5}$
Using $G^{2} = A H$ we get
$G = 6, A = \frac{15}{2}$ and $H = \frac{24}{5}$
Hence, $a + b = 15$ and $ab = 36$
$a = 12, b = 3$ or $a = 3, b = 12$
$∣ ∣ a^{2} - b^{2} ∣ ∣ = 135$ .