If the sum of $n$ terms of an A.P. is given by $S_{n} = a + b n + c n^{2}$ , where $a, b, c$ are independent of $n$ , then which of the following is (are) CORRECT?

a = 0

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Common difference of the A.P. must be

2 b

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Common difference of the A.P. must be

2 c

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First term of the A.P. is

b + c

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Solution

The correct option is D First term of the A.P. is $b + c$
Let $a^{'}$ be the first term and $d$ be the common difference.
$S_{n} = \frac{n}{2} [2 a^{'} + (n - 1) d] = a + b n + c n^{2} \Rightarrow n a^{'} + \frac{n (n - 1)}{2} d = a + b n + c n^{2} \Rightarrow (a^{'} - \frac{d}{2}) n + \frac{n^{2} d}{2} = a + b n + c n^{2}$
On comparing, we get
$a = 0, b = a^{'} - \frac{d}{2}$
and $c = \frac{d}{2} \Rightarrow d = 2 c$
Hence, first term $a^{'} = b + c$

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S_{n} = a + b n + c n^{2}

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S_{n} = a + b n + c n^{2}

	List 1		List 2
A.	If $a, b, c, d$ are in A.P., then (where $a, b, c & d$ $\in R^{+}$ )	1.	$a + d > b + c$
B.	If $a, b, c, d$ are in G.P, then (where $a, b, c & d$ $\in R^{+}$ )	2.	$b c > a d$
C.	The sum of $n$ terms of an A.P. is given by $S_{n} = b n + c n^{2}$ where $b$ and $c$ are independent of $n$ , then	3.	$T_{n} = b + (2 n - 1) c$
		4.	Common difference $(D) = (2 b - 4)$