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

A arithmetic progression is a progression for which the difference of each two consecutive terms is a constant and harmonic progression is a progression formed by taking the reciprocal of an arithmetic progression . Formula for finding $n$ th term in arithmetic progression is $n^{th}$ term $= a + (n - 1)d$ , where $a$ is the first term and $d$ is the common difference.

Complete step by step answer:
From the given data we know ${a_1} = {h_1} = 2$ and ${a_{10}} = {h_{10}} = 3$
Now we calculate $10$ th term in arithmetic progression
${a_{10}} = {a_1} + (10 - 1){d_1}$ , where ${a_1}$ is the first term and $d$ is common difference
Put the values ${a_1} = 2$ and ${a_{10}} = 3$ , calculate we get
$\Rightarrow 3 = 2 + 9{d_1}$
$\Rightarrow 3 - 2 = 9{d_1}$
$\Rightarrow 9{d_1} = 1$
Dividing both sides of above equation by $9$ , we get
$\Rightarrow {d_1} = \dfrac{1}{9}$
We know that the harmonic progression is reciprocal of arithmetic progression
Therefore $\dfrac{1}{{{h_1}}},\dfrac{1}{{{h_2}}},...,\dfrac{1}{{{h_{10}}}}$ are in arithmetic progression
Now we calculate $10$ th term in harmonic progression
$\dfrac{1}{{{h_{10}}}} = \dfrac{1}{{{h_1}}} + (10 - 1){d_2}$ , where $\dfrac{1}{{{h_1}}}$ is the first term and ${d_2}$ is common difference
We know ${h_1} = 2$ and ${h_{10}} = 3$ , then we get $\dfrac{1}{{{h_1}}} = \dfrac{1}{2}$ and $\dfrac{1}{{{h_{10}}}} = \dfrac{1}{3}$
Put the values $\dfrac{1}{{{h_1}}} = \dfrac{1}{2}$ and $\dfrac{1}{{{h_{10}}}} = \dfrac{1}{3}$ , calculate we get
$\dfrac{1}{3} = \dfrac{1}{2} + (10 - 1){d_2}$
$\Rightarrow 9{d_2} = \dfrac{1}{3} - \dfrac{1}{2}$
$\Rightarrow 9{d_2} = \dfrac{{2 - 3}}{6}$
$\Rightarrow 9{d_2} = - \dfrac{1}{6}$
Take cross multiplication and get
${d_2} = - \dfrac{1}{{54}}$
Now we find the value of ${a_4}$ in arithmetic progression , we get
${a_4} = {a_1} + (4 - 1){d_1}$
Put the value of ${a_1} = 2$ and ${d_1} = \dfrac{1}{9}$ , we get
$\Rightarrow {a_4} = 2 + 3 \times \dfrac{1}{9}$
$\Rightarrow {a_4} = 2 + \dfrac{1}{3}$
$\Rightarrow {a_4} = \dfrac{{6 + 1}}{3}$
$\Rightarrow {a_4} = \dfrac{7}{3}$
Now we find ${h_7}$ in harmonic progression i.e., $\dfrac{1}{{{h_7}}}$ in arithmetic progression .
$\dfrac{1}{{{h_7}}} = \dfrac{1}{{{h_1}}} + (7 - 1){d_2}$
Put the value of $\dfrac{1}{{{h_1}}} = \dfrac{1}{2}$ and ${d_2} = - \dfrac{1}{{54}}$ in above equation and get
$\Rightarrow \dfrac{1}{{{h_7}}} = \dfrac{1}{2} + 6 \times \left( { - \dfrac{1}{{54}}} \right)$
$\Rightarrow \dfrac{1}{{{h_7}}} = \dfrac{1}{2} - 6 \times \dfrac{1}{{54}}$
$\Rightarrow \dfrac{1}{{{h_7}}} = \dfrac{1}{2} - \dfrac{1}{9}$
Now we take $lcm(9,2) = 18$ and subtract , we get
$\Rightarrow \dfrac{1}{{{h_7}}} = \dfrac{{9 - 2}}{{18}}$
$\Rightarrow \dfrac{1}{{{h_7}}} = \dfrac{7}{{18}}$
We take in inversion of the above equation and get
$\Rightarrow {h_7} = \dfrac{{18}}{7}$
Now we have ${a_4} = \dfrac{7}{3}$ and ${h_7} = \dfrac{{18}}{7}$ ,
We now calculate the value of ${a_4}{h_7}$ is
${a_4}{h_7} = \dfrac{7}{3} \times \dfrac{{18}}{7}$
We know$\dfrac{7}{7} = 1$ , use this and get
$\Rightarrow {a_4}{h_7} = \dfrac{{18}}{3}$
$\Rightarrow {a_4}{h_7} = 6$
$\therefore$ The value of ${a_4}{h_7}$ is $6$. So, option (D) is correct.

Note:
When we take a function from left to the right side of equals then the sign will be changed and when we multiply any function then take special care about the sign . Rule of changes the signs are $( + ) \times ( - ) = ( - ),( + ) \times ( + ) = ( + ),( - ) \times ( - ) = ( + )$