Question 8

If the sum of n terms of an A.P. is (pn + qn²), where p and q are constants, find the common difference

Answer

It is known that,

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

According to the given condition,

\begin{align} \frac{n}{2}\left[2a + (n-1)d\right] = pn + qn^2 \end{align}

\begin{align} ⇒ \frac{n}{2}\left[2a + nd-d\right] = pn + qn^2 \end{align}

\begin{align} ⇒ na + n^2\frac{d}{2} - n.\frac{d}{2}= pn + qn^2 \end{align}

Comparing the coefficients of n² on both sides, we obtain

\begin{align} \frac{d}{2} = q \end{align}

\begin{align} \therefore d = 2q \end{align}

Thus, the common difference of the A.P. is 2q.

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