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Q.
The Sum of three consecutive number in HP is 37 and the sum of the reciprocals is $1 / 4$ then number are
Sequences and Series
Solution:
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}{1-12 d}+\frac{12}{1+12 d}=25 $
$\Rightarrow \frac{24}{25}=1-144 d^2 $
$ \Rightarrow d=\pm \frac{1}{60}$
Hence number are $\frac{1}{\frac{1}{12}-\frac{1}{60}}, 12, \frac{1}{\frac{1}{12}+\frac{1}{60}}=15,12,10$