Question

The $p^{th}$ term $T_p$ of H.P. is $q(p+q)$ and $q^{th}$ term $T_q$ is $p(p+q)$ where $p>1,q>1$ where $p\ne q$, then

• A. $T_{p+q}=pq$
• B. $T_{pq}=p+q$
• C. $T_{p+q}>T_{pq}$
• D. $T_{pq}>T_{p+q}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $T_{p+q}=pq$
B $T_{pq}=p+q$
C $T_{p+q}>T_{pq}$

Let the $1^{st}$ term be $a$ and the common difference of the corresponding AP be $d$

$T_p=\frac{1}{\frac{1}{a}+(p-1)d}=\frac{a}{1+(p-1)ad}=q(p+q)$

$\Rightarrow 1+(p-1)ad=\frac{a}{q(p+q)}$ ...(i)

$T_q=\frac{1}{\frac{1}{a}+(q-1)d}=\frac{a}{1+(q-1)ad}=p(p+q)$

$\Rightarrow 1+(q-1)ad=\frac{a}{p(p+q)}$ ...(ii)

Subtracting (i) and (ii), we get

$(q-p)ad=a.\left\{\frac{(q-p)(q+p)}{pq(p+q)}\right\}\Rightarrow d=\frac{1}{pq(p+q)}$...(iii)

Substituting (iii) in (ii), we get

$1+\frac{(q-1)a}{pq(p+q)}=\frac{a}{p(p+q)}\Rightarrow a=pq(p+q)$ ...(iv)

Thus, $T_{pq}=\frac{1}{\frac{1}{a}+(pq-1)d}=\frac{1}{\frac{1}{pq(p+q)}+\frac{pq-1}{pq(p+q)}}=p+q$

Again, $T_{p+q}=\frac{1}{\frac{1}{a}+(p+q-1)d}=\frac{1}{\frac{1}{pq(p+q)}+\frac{p+q-1}{pq(p+q)}}=pq$

Also, if $p$ and $q$ are more than $1$, where $p\ne q$ then $pq>p+q$

Thus, $T(p+q)>T\left(pq\right)$

Hence, (a), (b), (c) are correct.