Question

In an AP, if the pth term is \[\dfrac{1}{q}\] and qth term is \[\dfrac{1}{p}\]. Then, the sum of the first pq term is:

(a) \[\left( pq+1 \right)\]
(b) \[\dfrac{1}{2}\left( pq+1 \right)\]
(c) \[\dfrac{1}{2}\left( pq-1 \right)\]
(d) None of these

Answer 1

Hint: First of all, find the first term ‘a’’ and the common difference ‘d’ by using \[{{a}_{n}}=a+\left( n-1 \right)d\] and making two equations by the given information. Now use \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] and substitute the values of n, d, and a to given the sum of pq terms.

Complete step-by-step answer:

In this question, we are given that in an A.P, the pth term is \[\dfrac{1}{q}\] and qth term is \[\dfrac{1}{p}\]. We have to find the sum of the first pq terms. We know that the nth term of an A.P is given by

\[{{a}_{n}}=a+\left( n-1 \right)d.....\left( i \right)\]

where a is the first term of A.P, n is the number of terms in A.P and d is the common difference of A.P.

We are given that the pth term is \[\dfrac{1}{q}\]. So, by substituting \[{{a}_{n}}=\dfrac{1}{q}\] and n = p in equation (i), we get,

\[a+\left( p-1 \right)d=\dfrac{1}{q}....\left( ii \right)\]

Also, the qth term is \[\dfrac{1}{p}\]. So, we get,

\[a+\left( q-1 \right)d=\dfrac{1}{p}....\left( iii \right)\]

Now, by subtracting equation (iii) from equation (ii), we get,

\[\left[ a+\left( p-1 \right)d \right]-\left[ a+\left( q-1 \right)d \right]=\dfrac{1}{q}-\dfrac{1}{p}\]

By simplifying the above equation, we get,

\[\left( a+pd-d \right)-\left( a+qd-d \right)=\dfrac{1}{q}-\dfrac{1}{p}\]

\[\left( a-a \right)+\left( pd-qd \right)-d+d=\dfrac{1}{q}-\dfrac{1}{p}\]

By canceling the like terms, we get,

\[\left( p-q \right)d=\dfrac{\left( p-q \right)}{pq}\]

By dividing (p – q) on both the sides of the above equation, we get,

\[d=\dfrac{1}{pq}\]

Now, by substituting \[d=\dfrac{1}{pq}\] in equation (ii), we get,

\[a+\left( p-1 \right)d=\dfrac{1}{q}\]

\[\Rightarrow a+\left( p-1 \right)\dfrac{1}{pq}=\dfrac{1}{q}\]

\[\Rightarrow a+\dfrac{1}{q}-\dfrac{1}{pq}=\dfrac{1}{q}\]

By transferring the terms containing p and q to the RHS and canceling the like terms, we get,
\[a=\dfrac{1}{pq}\]

So, for this A.P, we get \[a=\dfrac{1}{pq}\] and \[d=\dfrac{1}{pq}\]

Now, we know that the sum of the n terms of A.P is given by

\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]

So, to find the sum of pq terms of this A.P, we substitute n = pq, \[a=\dfrac{1}{pq}\] and \[d=\dfrac{1}{pq}\]. We get,

\[{{S}_{pq}}=\dfrac{pq}{2}\left[ 2.\dfrac{1}{pq}+\left( pq-1 \right)\dfrac{1}{pq} \right]\]

\[{{S}_{pq}}=\dfrac{pq}{2}\left[ \dfrac{2}{pq}+\dfrac{pq}{pq}-\dfrac{1}{pq} \right]\]

\[{{S}_{pq}}=\dfrac{pq}{2}\left[ \dfrac{2}{pq}-\dfrac{1}{pq}+1 \right]\]

\[{{S}_{pq}}=\dfrac{pq}{2}\left[ \dfrac{1}{pq}+1 \right]\]

By taking pq inside the bracket, we get,

\[{{S}_{pq}}=\dfrac{1}{2}\left[ 1+pq \right]\]

So, we get the sum of pq terms as \[\dfrac{1}{2}\left( 1+pq \right)\].

Hence, option (b) is the right answer.

Note: In this question, students must take care while substituting the values of n because in each equation, its value is different while a and d remains constant for an A.P. Also, take proper care while calculating and taking out the common terms and canceling the terms as many students get confused between p and q.