Question

Let $a_1, a_2, \ldots ., a_{10}$ be in A.P. and $h_1, h_2, \ldots . ., h_{10}$ be in H.P. If $a_1=h_1=2$ and $a_{10}=h_{10}=3$ then $a _4 h _7$ is equal to

• A. 2
• B. 3
• C. 5
• D. 6

Answer 1

Answer: (D)

$a _1, a _2, \ldots ., a _{10} \text {, are in A.P. } $
$h _1, h _2, \ldots . ., h _{10} \text { are in H.P. } $
$\therefore \frac{1}{ h _1}, \frac{1}{ h _2}, \ldots . ., \frac{1}{ h _{10}} \text { are in A.P. } $
$\therefore a _{10}= a _1+9 d $
$3+2+9 d $
$\therefore d =\frac{1}{9} $
$\text { and } \frac{1}{ h _{10}}=\frac{1}{ h _1}+9 D ; \frac{1}{3}=\frac{1}{2}+9 D ; -\frac{1}{6}=9 D ; $
$ \therefore D =-\frac{1}{54} $
$\therefore a _4= a _1+3 d =2+\frac{1}{3}=\frac{7}{3}$
$\frac{1}{ h _7}=\frac{1}{ h _1}+6 D \Rightarrow \frac{1}{ h _7}=\frac{1}{2}+6 \times\left(-\frac{1}{54}\right)$
$ \Rightarrow \frac{1}{ h _7}=\frac{1}{2}-\frac{1}{9} \Rightarrow \frac{1}{ h _7}=\frac{7}{18} $
$\therefore h _7=\frac{18}{7} ;$
$\therefore a _4 h _7=\frac{7}{3} \times \frac{18}{7}=6$