Question

The pth term of an AP is 1/q and qth term is 1/p . The sum of the pqth term is what​

Answer 1

Solution:

Given:

• pth term of an AP = 1/q
• qth term of an AP = 1/p

To Find:

• Sum of the p and qth term.

Now, let first term = a and common difference = d.

Now, according to question,

$\implies \sf a_{p}=\dfrac{1}{q}=a+(p-1)d=\dfrac{1}{q}\;\;\;\;........(1)\\ \\ \\ \implies \sf a_{q}=\dfrac{1}{p}=a+(q-1)d=\dfrac{1}{p}\;\;\;\;........(2)\\ \\ \\ \underline{\bf Now, \;equation\;(1)-equation\;(2)}\\ \\ \\ \implies \sf (p-1)d-(q-1)d=\dfrac{1}{q} -\dfrac{1}{p}\\ \\ \\ \implies \sf d[p-1-q+1]=\dfrac{p-q}{pq}\\ \\ \\ \implies d=\dfrac{1}{qp}$

$\underline{\bf Now,\;put\;the\;value\;of\;d\;in\;equation\;(1)}\\ \\ \\ \implies \sf a+(p-1)\Bigg(\dfrac{1}{pq}\Bigg)=\dfrac{1}{q}\\ \\ \\ \implies \sf a+\Bigg(\dfrac{p}{pq}-\dfrac{1}{pq}\Bigg)=1\\ \\ \\ \implies \sf a=\dfrac{1}{pq}\\ \\ \\ \underline{\bf Now,\;by\;S_{n}\;formula,} \\ \\ \\ \implies \sf S_{n}=\dfrac{n}{2}[2a+(n-1)d]\\ \\ \\ \implies S_{pq}=\dfrac{pq}{2}\Bigg[\dfrac{2}{pq}+(pq-1)\Bigg(\dfrac{1}{pq}\Bigg)\Bigg]$

$\implies \sf S_{pq}=\dfrac{pq}{2}\Bigg[\dfrac{2}{pq}+\dfrac{pq}{pq}-\dfrac{1}{pq}\Bigg]\\ \\ \\ \implies \sf S_{pq}=\dfrac{pq}{2}\Bigg[\dfrac{1}{pq}+1\Bigg]\\ \\ \\ \implies \sf S_{pq}=\dfrac{pq}{2}\Bigg[\dfrac{1+pq}{pq}\Bigg]\\ \\ \\ \implies {\boxed{\sf S_{pq}=\dfrac{1}{2}(1+pq)}}$

Answer 2

plz refer to this attachment