Mathematics - Sequence & Series Question with Solution | TestHub

Let $d$ be the common difference of the AP. Then, $\begin{array}{rl}& a_{10}=3\Rightarrow a_1+9d=3\\ \Rightarrow & 2+9d=3\Rightarrow d=\frac{1}{9}\\ \therefore & a_4=a_1+3d=2+\frac{1}{3}=\frac{7}{3}\end{array}$ Let $D$ be the common difference of $\frac{1}{h_1},\frac{1}{h_2},\dots .-\frac{1}{h_{10}}$. Then, $h_{10}=3$ $\begin{array}{ll}\Rightarrow & \frac{1}{h_{10}}=\frac{1}{3}\Rightarrow \frac{1}{2}+9D=\frac{1}{3}\\ \Rightarrow & 9D=-\frac{1}{6}\Rightarrow D=-\frac{1}{54}\\ \therefore & \frac{1}{h_7}=\frac{1}{h_1}+6D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\\ \Rightarrow & h_7=\frac{18}{7}\\ \therefore & a_4{h}_7=\frac{7}{3}\times \frac{18}{7}=6\end{array}$