Question

If $a_1,a_2,...a_{10}$be in AP, $\frac{1}{h_1},\frac{1}{h_2},...\frac{1}{h_{10}}$ be in AP. $a_1=h_1=2$and $a_{10}=h_{10}=3$ then $a_4{h}_7=?$

Answer 1

Explanation for the correct option:

Step 1. Find the Value of ${\mathit{a}}_{\mathbf{4}}\mathbf{}{\mathit{h}}_{\mathbf{7}}$

Given $a_1,a_2,..a_{10}$and $h_1,h_2,..h_{10}$ are in

$$\begin{gathered} a_1=h_1=2 \\ {a}_{10}=h_{10}=3 \end{gathered}$$

Step 2. Finding $a_4$

$a_{10}=a_1+(10-1)d_1$

$\Rightarrow$$3=2+9d_1$

$\Rightarrow$$d_1=\frac{1}{9}$

$\begin{array}{rcl}{a}_4& =& a_1+3d_1\\ & =& 2+3\times \frac{1}{9}\\ & =& 2+\frac{1}{3}\\ & =& \frac{7}{3}\end{array}$

Step 3. Finding $h_7$

$\frac{1}{h_{10}}=\frac{1}{h_1}+9d_2$

$\Rightarrow$ $\frac{1}{3}=\frac{1}{2}+9d_2$

$\Rightarrow$ $d_2=-\frac{1}{54}$

$\Rightarrow$ $\frac{1}{h_7}=\frac{1}{h_7}+6d_2$

$\Rightarrow$ $\frac{1}{h_7}=\frac{1}{2}+6\times (-\frac{1}{54})$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{h_7}& =& \frac{1}{2}-\frac{1}{9}\\ & =& \frac{7}{18}\end{array}$

$\Rightarrow$ $h_7=\frac{18}{7}$

Step 4. Multiply $a_4$ and $h_7$

$\begin{array}{rcl}{a}_4{h}_7& =& \frac{7}{3}\times \frac{18}{7}\\ & =& 6\end{array}$

Hence, the value of $a_4{h}_7$ is $\mathbf{6}$