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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Welcome to the NCERT Solutions for Class 11 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 8: If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common differe....
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