Question

If we are given the series ${a_{1,}}{a_2}.....{a_{10}}$be in AP, ${h_{1,}}{h_2}.....{h_{10}}$be in HP. ${a_1} = {h_1} = 2,{a_{10}} = {h_{10}} = 3$, then ${a_4}{h_7} = $
1. 2
2. 3
3. 5
4. 6

Answer 1

Hint: The first term and the tenth term of the Arithmetic Progression (AP) and the Harmonic Progression (HP) is given. From those terms, find out the common differences for each of them. Then find out the terms a4 and h7. Then multiply them, which gives us the answer.
Formulae used in the problem are:
For Arithmetic Progression:
${a_n} = {a_1} + \left( {n - 1} \right)d$
Where,
${a_n} - {n^{th}}$ term in the sequence
${a_1} - {1^{st}}$ term in the sequence
$d - $common difference between the terms
For Harmonic Progression:
$\dfrac{1}{{{h_n}}} = \dfrac{1}{{{h_1}}} + \left( {n - 1} \right)d$
Where,
${h_n} - {n^{th}}$term in the sequence
${h_1} - {1^{st}}$term in the sequence
$d - $common difference between the terms

Complete step-by-step solution:
For the Arithmetic Progression, it is given that
 $\eqalign{
  & {a_1} = 2 \cr
  & {a_{10}} = 3 \cr} $
Let us find the common difference, d
${a_n} = {a_1} + \left( {n - 1} \right)d$
Substituting the given,
$\eqalign{
  & \Rightarrow {a_{10}} = {a_1} + \left( {10 - 1} \right)d \cr
  & \Rightarrow 3 = 2 + 9d \cr
  & \Rightarrow 9d = 1 \cr
  & \Rightarrow d = \dfrac{1}{9} \cr} $
Now we can find the ${4^{th}}$term, ${a_4}$
$\eqalign{
  & {a_4} = {a_1} + \left( {4 - 1} \right)d \cr
  & \Rightarrow {a_4} = 2 + 3\left( {\dfrac{1}{9}} \right) \cr
  & \Rightarrow {a_4} = \dfrac{7}{3} \cr} $
Now, let us move on to Harmonic Progression. It is given that,
$\eqalign{
  & {h_1} = 2 \cr
  & {h_{10}} = 3 \cr} $
Let us find the common difference, d
$\eqalign{
  & \Rightarrow \dfrac{1}{{{h_n}}} = \dfrac{1}{{{h_1}}} + \left( {n - 1} \right)d \cr
  & \Rightarrow \dfrac{1}{{{h_{10}}}} = \dfrac{1}{{{h_1}}} + \left( {10 - 1} \right)d \cr
  & \Rightarrow \dfrac{1}{3} = \dfrac{1}{2} + 9d \cr
  & \Rightarrow \dfrac{1}{3} - \dfrac{1}{2} = 9d \cr
  & \Rightarrow d = \dfrac{{ - 1}}{{54}} \cr} $
Now, let us find out ${h_7}$
$\eqalign{
  & \Rightarrow \dfrac{1}{{{h_7}}} = \dfrac{1}{{{h_1}}} + \left( {7 - 1} \right)d \cr
  & \Rightarrow \dfrac{1}{{{h_7}}} = \dfrac{1}{2} + 6\left( {\dfrac{{ - 1}}{{54}}} \right) \cr
  & \Rightarrow \dfrac{1}{{{h_7}}} = \dfrac{7}{{18}} \cr
  & \Rightarrow {h_7} = \dfrac{{18}}{7} \cr} $
Since we know both the required values, we can multiply these to get the final answer.
$\eqalign{
  & \therefore {a_4}{h_7} = \dfrac{7}{3} \times \dfrac{{18}}{7} \cr
  & \Rightarrow {a_4}{h_7} = 6 \cr} $
The final answer is $6$.
Hence, option (4) is the correct answer.

Note: Since the first and last terms for both the AP and HP are same, be careful while substituting them. The AP and the HP have very similar formulae, except, the HP is the reciprocal of AP. Since HP is the reciprocal, the last step will have to be reciprocated to get the correct value.