Question

Let $a_1,a_2................,a_{10}$ be in AP and $h_1,h_2,........h_{10}$ be in HP. If $a_1=h_1=2$ and $a_{10}=3$, then a$a_4{h}_7$ is

• A. 2
• B. 3
• C. 5
• D. 6

Answer 1

The correct option is D 6

$a_{10}=3\Rightarrow \Rightarrow a_{10}=3\Rightarrow a_1+9d=3\Rightarrow 2+9d=3\Rightarrow d=\frac{1}{9}\therefore a_4=a_1+3d=2+\frac{1}{3}=\frac{7}{3}$
$h_{10}=3\Rightarrow \frac{1}{h_{10}}=\frac{1}{3}\Rightarrow D=-\frac{1}{54}\therefore \frac{1}{h_7}=\frac{1}{h_1}+6D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\Rightarrow h_7=\frac{18}{7}Hence,a_4{h}_7=\frac{7}{3}\times \frac{18}{7}=6$