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 $a_{10}=h_{10}=3$, then $a_4{h}_7$ is

• A. $2$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is D

$6$

Explanation for correct option

Step 1:Determine value of $a_4$

An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant.

$$\begin{gathered} \Rightarrow a_n=a_1+\left(n-1\right)d_a \\ \Rightarrow a_{10}=a_1+\left(10-1\right)d_a \\ \Rightarrow 3=2+9d_a[\because a_{10}=3,a_1=2\mathrm{given}] \\ \Rightarrow d_a=\frac{1}{9} \end{gathered}$$

Again we have,

$$\begin{gathered} a_4=a_1+\left(4-1\right)d_a \\ =2+3\times \frac{1}{9} \\ \Rightarrow a_4=\frac{7}{3}...\left(i\right) \end{gathered}$$

Step 2: Determine value of $h_7$

A Harmonic Progression (HP) is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progression that does not contain 0.

$h_1=2=\frac{1}{a_1}$ [Given]

$h_{10}=3=\frac{1}{a_{10}}$ [Given]

$$\begin{gathered} \Rightarrow h_n=\frac{1}{a_1+\left(n-1\right)d_h} \\ \Rightarrow h_{10}=\frac{1}{a_1+9d_h} \\ \Rightarrow 3=\frac{1}{{\displaystyle \frac{1}{2}}+9d_h}\left[\because \frac{1}{a_1}=2\right] \\ \Rightarrow 3=\frac{2}{{\displaystyle 1+18d_h}} \\ \Rightarrow 2=3\left(1+18d_h\right) \\ \Rightarrow -1=54d_h \\ \Rightarrow d_h=\frac{-1}{54} \end{gathered}$$

Therefore $h_7$, can be given as :

$$\begin{gathered} h_7=\frac{1}{a_1+\left(7-1\right)d_h} \\ =\frac{1}{{\displaystyle \frac{1}{2}}+6\left({\displaystyle \frac{-1}{54}}\right)} \\ =\frac{1}{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{9}}} \\ {h}_7=\frac{18}{7}...\left(ii\right) \end{gathered}$$

Step 3: Calculate $a_4{h}_7$:

From $\left(i\right)$ and $\left(ii\right)$, we have

$$\begin{gathered} a_4{h}_7=\frac{7}{3}\times \frac{18}{7} \\ =\frac{18}{3} \\ {a}_4{h}_7=6 \end{gathered}$$

Hence, option (D) is the correct answer.