Question

Let ${a_1},{a_2},...,{a_{10}}$ be in A.P and ${h_1},{h_2},...,{h_{10}}$ be in H.P. 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

Here we use the formula for nth term of an AP and find the value of common difference in AP. Using the value of first term and common difference we find the value of ${a_4}$. Now we write the terms of H.P in terms of an AP and find the common difference using the formula of the nth term of an AP. Use the value of common difference and first term to find the value of ${h_7}$.

• An arithmetic progression is a sequence of terms having common differences between them. If a is the first term of an AP, d is the common difference, then the nth term of an AP can be found as ${a_n} = a + (n - 1)d$.
• A harmonic progression is a sequence of real numbers which is made by taking reciprocal terms of an AP. If we have an AP with terms ${a_1},{a_2}......$ then the HP formed from the terms of the AP is $\dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},.....$

Complete step-by-step answer:
We are given${a_1},{a_2},...,{a_{10}}$is an AP.
Then the first term is ${a_1}$
We know the value of${a_1} = 2$.
Let the common difference between the terms be$d$
Then the nth term of the AP can be calculated using the formula${a_n} = a + (n - 1)d$
Now we are given the value of 10th term of AP, ${a_{10}} = 3$
Substitute the value of n as 10 and a as 2
$\Rightarrow 3 = 2 + (10 - 1)d$
$\Rightarrow 3 = 2 + 9d$
Shift all constant terms to one side of the equation.
$\Rightarrow 3 - 2 = 9d$
$\Rightarrow 1 = 9d$
Divide both sides of the equation by 9
$\Rightarrow \dfrac{1}{9} = \dfrac{{9d}}{9}$
Cancel same factor from numerator and denominator.
$\Rightarrow d = \dfrac{1}{9}$
Now we find the value of ${a_4}$using the formula${a_n} = a + (n - 1)d$
$\Rightarrow {a_4} = a + (4 - 1)d$
$\Rightarrow {a_4} = a + 3d$
Substitute the value of a as 2, d as $\dfrac{1}{9}$
$\Rightarrow {a_4} = 2 + 3 \times \dfrac{1}{9}$
Cancel same factor from numerator and denominator in RHS
$\Rightarrow {a_4} = 2 + \dfrac{1}{3}$
Take LCM
$\Rightarrow {a_4} = \dfrac{{6 + 1}}{3}$
$\Rightarrow {a_4} = \dfrac{7}{3}$ … (1)
Now we are given the terms${h_1},{h_2},...,{h_{10}}$are in HP.
We have ${h_1} = 2$and ${h_{10}} = 3$
We know the terms of a harmonic progression are the reciprocal of terms of an arithmetic progression.
Then the AP formed by the terms of the given HP is $\dfrac{1}{{{h_1}}},\dfrac{1}{{{h_2}}},.....,\dfrac{1}{{{h_{10}}}}$
Since the terms are in AP, let the common difference be$d'$.
Then the nth term of this AP can be found using the formula${a_n}' = a' + (n - 1)d'$
We have the value of $a' = \dfrac{1}{{{h_1}}}$
Substitute the value of ${h_1} = 2$
$\Rightarrow a' = \dfrac{1}{2}$
Similarly, we have the 10th term of this AP as ${a_{10}}' = \dfrac{1}{{{h_{10}}}}$
Substitute the value of${h_{10}} = 3$
$\Rightarrow {a_{10}}' = \dfrac{1}{3}$
Substitute the values of$a' = \dfrac{1}{2}$and${a_{10}}' = \dfrac{1}{3}$in the general formula of nth term, taking n as 10.
$\Rightarrow {a_{10}}' = a' + (10 - 1)d'$
$\Rightarrow \dfrac{1}{3} = \dfrac{1}{2} + 9d'$
Shift the constant values to one side of the equation.
$\Rightarrow \dfrac{1}{3} - \dfrac{1}{2} = 9d'$
Take LCM
$\Rightarrow \dfrac{{2 - 3}}{{3 \times 2}} = 9d'$
$\Rightarrow \dfrac{{ - 1}}{6} = 9d'$
Divide both sides by 9
$\Rightarrow \dfrac{{ - 1}}{{6 \times 9}} = \dfrac{{9d'}}{9}$
Cancel the same terms from numerator and denominator.
$\Rightarrow \dfrac{{ - 1}}{{54}} = d'$
Now we find the value of ${a_7}$using the formula${a_n}' = a' + (n - 1)d'$
Substitute the value of a’ as $\dfrac{1}{2}$ and d’ as $- \dfrac{1}{{54}}$and n as 7
$\Rightarrow {a_7} = \dfrac{1}{2} + (7 - 1)\left( {\dfrac{{ - 1}}{{54}}} \right)$
$\Rightarrow {a_7} = \dfrac{1}{2} + 6 \times \dfrac{{ - 1}}{{54}}$
Cancel the same terms from numerator and denominator.
$\Rightarrow {a_7} = \dfrac{1}{2} - \dfrac{1}{9}$
Take LCM
$\Rightarrow {a_7} = \dfrac{{9 - 2}}{{2 \times 9}}$
$\Rightarrow {a_7} = \dfrac{7}{{18}}$
We know that value of term in HP is the reciprocal of term in AP
Then, ${h_7} = \dfrac{1}{{{a_7}}}$
Substituting the value of ${a_7} = \dfrac{7}{{18}}$
$\Rightarrow {h_7} = \dfrac{{18}}{7}$ ... (2)
Then value of ${a_4}{h_7}$can be found by multiplying the values from equation (1) and (2)
$\Rightarrow {a_4}{h_7} = \dfrac{7}{3} \times \dfrac{{18}}{7}$
Cancel the same terms from numerator and denominator.
$\Rightarrow {a_4}{h_7} = 6$

Therefore, the correct option is D.

Note: Students many times make mistakes while calculating the common difference of HP. They take the given terms of HP as the terms of the corresponding AP. Keep in mind the terms of HP will look like $\dfrac{1}{a},\dfrac{1}{{a + d}},\dfrac{1}{{a + 2d}},.......$ where the terms in the denominator form an AP.
Always change signs when shifting a value from one side of the equation to another side.