Question

In H.P, ${T_3} = \dfrac{1}{7}$ and ${T_7} = \dfrac{1}{5}$. Find ${T_{15}}$.

Answer 1

Hint: This question is based on H.P, about which we know that $HP = \dfrac{1}{{AP}}$, where H.P is Harmonic progression and A.P is Arithmetic progression. So, we can convert the terms of H.P into terms of A.P then by that we can find the common difference of A.P and first term of A.P, then we can find its ${15^{th}}$term and then by reciprocating it we will get our answer.

Complete step-by-step answer:
In this question first of we will start solving this question by enlisting the given information.
So, it is given in the question that in a H.P, ${T_3} = \dfrac{1}{7}$ and ${T_7} = \dfrac{1}{5}$.
So, now we will convert this H.P series into an A.P series, because it is easier to solve an A.P series directly as compared to solving a H.P series as we know that $HP = \dfrac{1}{{AP}}$, where H.P is Harmonic progression and A.P is Arithmetic progression.
So, let us assume that ${t_i}$ is a general term of A.P like ${T_i}$ is of H.P, where $i$ is the ${i^{th}}$ no. of term.
Now, as$HP = \dfrac{1}{{AP}}$, we can say that ${T_i} = \dfrac{1}{{{t_i}}}$,
$ \Rightarrow {t_i} = \dfrac{1}{{{T_i}}}$
$ \Rightarrow {t_3} = \dfrac{1}{{{T_3}}} = \dfrac{1}{{\dfrac{1}{7}}}$ (As it is given that ${T_3} = \dfrac{1}{7}$)
$ \Rightarrow {t_3} = 7$
Also, ${t_7} = \dfrac{1}{{{T_7}}}$
$ \Rightarrow {t_7} = \dfrac{1}{{{T_7}}} = \dfrac{1}{{\dfrac{1}{5}}}$ (As it is given that ${T_7} = \dfrac{1}{5}$)
${t_7} = 5$
Also as they are in A.P we can form a general term for this A.P series which will be
${t_i} = a + \left( {i - 1} \right)d$
where $a = $ First term of A.P
$d = $ Common difference of this A.P series
And $i$ can be any no. depending upon the no. of that term.
So, if we consider ${t_3} = 7$
$ \Rightarrow {t_3} = 7 = a + \left( {3 - 1} \right)d$
$ \Rightarrow a + 2d = 7$ ……..(i)
Now, considering ${t_7} = 5$
$ \Rightarrow {t_7} = 5 = a + \left( {7 - 1} \right)d$
$ \Rightarrow a + 6d = 5$ ………..(ii)
Now subtracting equation (i) from (ii) we get,
$
   \Rightarrow {t_3} - {t_7} = 7 - 5 \\
   \Rightarrow \left( {a + 2d} \right) - \left( {a + 6d} \right) = 7 - 5 \\
   \Rightarrow a + 2d - a - 6d = 2 \\
   \Rightarrow - 4d = 2 \\
   \Rightarrow d = - \dfrac{1}{2} \\
 $
Now putting this value of $d$ in equation (i) to find the value of $a$,
$ \Rightarrow a + 2d = 7$
$
   \Rightarrow a + 2\left( { - \dfrac{1}{2}} \right) = 7 \\
   \Rightarrow a - 1 = 7 \\
   \Rightarrow a = 8 \\
 $
Now as per asked in the question we need to find the value of ${T_{15}}$, which is nothing but the reciprocal of ${t_{15}}$.
So, if we find ${t_{15}}$ and then by reciprocating it we will get our answer
So, as we know through the general term of A.P that ${t_{15}}$ will be equal to
${t_{15}} = a + \left( {15 - 1} \right)d$
Now, as we found above that $a = 8,d = - \dfrac{1}{2}$, so by putting their values we get
$
   \Rightarrow {t_{15}} = 8 + \left( {15 - 1} \right)\left( { - \dfrac{1}{2}} \right) \\
   \Rightarrow {t_{15}} = 8 + \left( {14} \right)\left( { - \dfrac{1}{2}} \right) \\
   \Rightarrow {t_{15}} = 8 + \left( { - 7} \right) \\
   \Rightarrow {t_{15}} = 1 \\
 $
As, we proved above that $HP = \dfrac{1}{{AP}}$,
$ \Rightarrow $${T_{15}} = \dfrac{1}{{{t_{15}}}}$
$ \Rightarrow $${T_{15}} = \dfrac{1}{{{t_{15}}}} = \dfrac{1}{1}$
$ \Rightarrow {T_{15}} = 1$

Note: This question can also be done by simply using the general term formula ${T_i} = \dfrac{1}{{a + \left( {i - 1} \right)d}}$ of H.P series $\dfrac{1}{a},\dfrac{1}{{a + d}},\dfrac{1}{{a + 2d}}$. In this case we can find the value of $a\& d$ and by that we can find the value of ${T_{15}}$ by putting $i = 15$ in the general term formula of H.P, in this case we don’t need to convert the H.P into A.P moreover both the formulas are correct.