Question

The value of x + y + z is 15 if a, x, y, b are in A.P. while the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}is\frac{5}{8}$ if a, x, y, z b are H.P., then $a^2+b^2=$

• A. 48
• B. 50
• C. 52
• D. 60

Answer 1

The correct option is C

52

a, x, y, b are in A.P $\Rightarrow \frac{a+b}{2}$

x + z = a + d + b – d = a + b, where d is the common difference

$\therefore x+y+z=15\Rightarrow \frac{a+b}{2}+(a+b)=15\Rightarrow a+b=10$

a, x, y, z, b are in H.P $\Rightarrow \frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{b}$ are in A.P

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b})+(\frac{1}{a}+\frac{1}{b})=\frac{3}{2}(\frac{1}{a}+\frac{1}{b})=\frac{3}{2}\frac{(a+b)}{ab}=\frac{5}{8}$,given

$\therefore ab=\frac{8}{5}\times \frac{3}{2}\times 10=24$ [using(1)]

Using (1) and (2) $\Rightarrow$ (a, b) = (4, 6) $\Rightarrow a^2+b^2=52$