Question

If $a,b,c(a<b<c)$ are in H.P. such that $a+b+c=37$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$, then which of the following is/are correct?

• A. $a,b,c-1$ are in A.P.
• B. $a-4,b,c+9$ are in G.P.
• C. A.M. of $a,b$ is $15$
• D. $\frac{a}{2},\frac{b}{3},\frac{c}{5}$ are sides of a right angled triangle

Answer 1

Answer: (A), (B), (D)

The correct options are
A $a,b,c-1$ are in A.P.
B $a-4,b,c+9$ are in G.P.
D $\frac{a}{2},\frac{b}{3},\frac{c}{5}$ are sides of a right angled triangle

$a,b,c$ are in H.P.
$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{4}\Rightarrow \frac{3}{b}=\frac{1}{4}(\because \frac{1}{a}+\frac{1}{c}=\frac{2}{b})\Rightarrow b=12$
Now,
$a+b+c=37\Rightarrow a+c=25\cdots \left(1\right)$
We know that,
$b=\frac{2ac}{a+c}\Rightarrow 12=\frac{2ac}{25}\Rightarrow ac=150$
Using equation $\left(1\right)$,
$\Rightarrow a(25-a)=150\Rightarrow a^2-25a+150=0\Rightarrow (a-10)(a-15)=0\Rightarrow a=10(\because a<b<c)\Rightarrow c=15$

$a,b,c-1$
$=10,12,14$ are in A.P.

$a-4,b,c+9$
$=6,12,24$ are in G.P.

A.M. of $a,b$
$=\frac{a+b}{2}=11$

$\frac{a}{2},\frac{b}{3},\frac{c}{5}$
$=5,4,3$ are sides of a right angled triangle.