Question

If the (2p)th term of a H.P is q and the (2q)th term is p, then the (2(p + q)th term is

• A. $\frac{pq}{2(p+q)}$
• B. $\frac{2pq}{p+q}$
• C. $\frac{pq}{p+q}$
• D. $\frac{p+q}{pq}$

Answer 1

The correct option is C $\frac{pq}{p+q}$

Let first term be a and the common difference be d.

Given, $2p^{th}$ term of H.P. is q so, $2p^{th}$ of that A.P. is $\frac{1}{q}$ so, $t_{2p}=a+(2p-1)d\Rightarrow \frac{1}{q}=a+(2p-1)d....\left(1\right)$

Similarly, $2q^{th}$ term of H.P. is p so, $2q^{th}$ of that A.P. is $\frac{1}{p}$ so, $t_{2q}=a+(2q-1)d\Rightarrow \frac{1}{p}=a+(2q-1)d....\left(2\right)$

Subtracting equation 1 from 2,

$\frac{1}{q}-\frac{1}{p}=a+(2p-1)d-a-(2q-1)d\Rightarrow \frac{p-q}{pq}=(2p-1-2q+1)d\Rightarrow d=\frac{p-q}{2pq(p-q)}\Rightarrow d=\frac{1}{2pq}$

Putting the value of d in equation 1,

$\frac{1}{q}=a+(2p-1)\times \frac{1}{2pq}\Rightarrow \frac{1}{q}=a+\frac{1}{q}-\frac{1}{2pq}\Rightarrow a=\frac{1}{2pq}$

Thus, $2(p+q{)}^{th}$ term of A.P. $=t_{2(p+q)}=\left[\frac{1}{2pq}\right]+\left[2\right(p+q)-1]\times \frac{1}{2pq}=\frac{p+q}{pq}$

So, $2(p+q{)}^{th}$ term of H.P. $=\frac{pq}{p+q}$