Question

If ${{p}^{th}}$ term and ${{q}^{th}}$ term of an A.P. are \[\dfrac{1}{qr}\] and \[\dfrac{1}{pr}\] respectively, then ${{r}^{th}}$ term of the A.P. is
A. \[\dfrac{1}{pqr}\]
B. 1
C. \[\dfrac{1}{pq}\]
D. \[pq\]

Answer 1

Hint: We assume the general terms of an arithmetic series. We find the formula for ${{t}_{n}}$, the ${{n}^{th}}$ term of the series. From that we express the ${{p}^{th}}$, ${{q}^{th}}$, ${{r}^{th}}$ term and use the given relations to find the variables. We solve it and use that to find the ${{r}^{th}}$ term of the A.P..

Complete step by step answer:
We express the arithmetic sequence in its general form. We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series. The first term be ${{t}_{1}}$ and the common difference be $d$ where $d={{t}_{2}}-{{t}_{1}}={{t}_{3}}-{{t}_{2}}={{t}_{4}}-{{t}_{3}}$. We can express the general term ${{t}_{n}}$ based on the first term and the common difference. The formula being ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d$. Therefore, the ${{p}^{th}}$, ${{q}^{th}}$, ${{r}^{th}}$ terms will be ${{t}_{p}}={{t}_{1}}+\left( p-1 \right)d$, ${{t}_{q}}={{t}_{1}}+\left( q-1 \right)d$, ${{t}_{r}}={{t}_{1}}+\left( r-1 \right)d$ respectively.

It is given that ${{p}^{th}}$ term and ${{q}^{th}}$ term of the A.P. are \[\dfrac{1}{qr}\] and \[\dfrac{1}{pr}\] respectively. Therefore, ${{t}_{p}}={{t}_{1}}+\left( p-1 \right)d=\dfrac{1}{qr}$ and ${{t}_{q}}={{t}_{1}}+\left( q-1 \right)d=\dfrac{1}{pr}$. We subtract those equations,
\[{{t}_{q}}-{{t}_{p}}=\left[ {{t}_{1}}+\left( q-1 \right)d \right]-\left[ {{t}_{1}}+\left( p-1 \right)d \right] \\
\Rightarrow {{t}_{q}}-{{t}_{p}} =\dfrac{1}{pr}-\dfrac{1}{qr} \\
\Rightarrow \left( q-p \right)d=\dfrac{1}{r}\left( \dfrac{1}{p}-\dfrac{1}{q} \right)=\dfrac{q-p}{pqr} \\
\Rightarrow d=\dfrac{1}{pqr} \\ \]
Putting the value of \[d=\dfrac{1}{pqr}\] in ${{t}_{p}}={{t}_{1}}+\left( p-1 \right)d=\dfrac{1}{qr}$, we get
${{t}_{p}}={{t}_{1}}+\dfrac{\left( p-1 \right)}{pqr}=\dfrac{1}{qr} \\
\Rightarrow {{t}_{1}}=\dfrac{1}{qr}-\dfrac{\left( p-1 \right)}{pqr}=\dfrac{1}{pqr} \\ $
Now we put ${{t}_{1}}=d=\dfrac{1}{pqr}$ in ${{t}_{r}}={{t}_{1}}+\left( r-1 \right)d$ to get ${{r}^{th}}$ term of the A.P. So,
${{t}_{r}}=\dfrac{1}{pqr}+\left( r-1 \right)\dfrac{1}{pqr} \\
\Rightarrow {{t}_{r}} =\dfrac{1+r-1}{pqr} \\
\therefore {{t}_{r}} =\dfrac{1}{pq}$

Hence, the correct option is C.

Note: The sequence is an increasing sequence where the common difference is a positive number. We have to take the values $p\ne 0;q\ne 0;r\ne 0$ as then the A.P. becomes obsolete. The common difference will never be calculated according to the difference of greater number from the lesser number.