Question 5

In an A.P., if pth term is \begin{align} \frac{1}{q} \; and \;qth\; term \; is\; \frac{1}{p}\end{align} , prove that the sum of first pq terms is \begin{align} \frac{1}{2}(pq+1) \; where \; p ≠ q \end{align}

Answer

It is known that the general term of an A.P. is an = a + (n – 1)d

∴ According to the given information,

\begin{align}p^{th} \; term= a_p=a+(p-1)d=\frac{1}{q} \;\; ...(1)\end{align}

\begin{align}q^{th} \; term= a_q=a+(q-1)d=\frac{1}{p} \;\; ...(2)\end{align}

Subtracting (2) from (1), we obtain

\begin{align} (p-1)d - (q-1)d=\frac{1}{q}-\frac{1}{p} \end{align}

\begin{align} ⇒(p-1-q+1)d = \frac{p-q}{pq} \end{align}

\begin{align} ⇒(p-q)d = \frac{p-q}{pq} \end{align}

\begin{align} ⇒d = \frac{1}{pq} \end{align}

Putting the value of d in (1), we obtain

\begin{align} a + (p-1)\frac{1}{pq}= \frac{1}{q}\end{align}

\begin{align} ⇒a = \frac{1}{q}-\frac{1}{q}+\frac{1}{pq}=\frac{1}{pq} \end{align}

\begin{align} \therefore S_{pq} = \frac{pq}{2}\left[2a + (pq-1)d\right]\end{align}

\begin{align} = \frac{pq}{2}\left[\frac{2}{pq} + (pq-1)\frac{1}{pq}\right]\end{align}

\begin{align} = 1 + \frac{1}{2}(pq-1)\end{align}

\begin{align} =\frac{1}{2}pq + 1 - \frac{1}{2}=\frac{1}{2}pq + \frac{1}{2}\end{align}

\begin{align} =\frac{1}{2}(pq+1) \end{align}

Thus, the sum of first pq terms of the A.P. is \begin{align} =\frac{1}{2}(pq+1). \end{align}

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