Question

The Sum of three consecutive number in HP is 37 and the sum of the reciprocals is $1/4$ then number are

• A. $9,10,12$
• B. $9,10,13$
• C. $15,12,10$
• D. $14,11,9$

Answer 1

Answer: (C)

let three numbers in HP are $\frac{1}{a-d},\frac{1}{a},\frac{1}{a+d}$
$\Rightarrow \frac{1}{a-d}+\frac{1}{a}+\frac{1}{a+d}=37$......(i)
and $a-d+a+a+d=\frac{1}{4}$
$\Rightarrow a=\frac{1}{12}$
put the value of a in (i) $\Rightarrow \frac{12}{1-12d}+\frac{12}{1+12d}=25$
$\Rightarrow \frac{24}{25}=1-144d^2$
$\Rightarrow d=\pm \frac{1}{60}$
Hence number are $\frac{1}{\frac{1}{12}-\frac{1}{60}},12,\frac{1}{\frac{1}{12}+\frac{1}{60}}=15,12,10$