Question

Let $a_1,a_2,\cdots a_{10}$ be in A.P and $h_1,h_2,\cdots h_{10}$ be in H.P. If $a_1=h_1=2$ and $a_{10}=h_{10}=3$
then $a_4{h}_7$ is :

• A. 2
• B. 3
• C. 5
• D. 6

Answer 1

The correct option is D 6

$a_1=2$

$a_{10}=a+9d=3$

$\Rightarrow d=\frac{1}{9}$

$\Rightarrow a_4=\frac{7}{3}$

$h_1=\frac{1}{a_1}=2$

$h_{10}=\frac{1}{a_{10}}=3$

$d=-\frac{1}{54}$

$h_7=\frac{1}{a+6d}=\frac{18}{7}$

$a_4\times h_7=\frac{7}{3}\times \frac{18}{7}=6$