If 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-

a = 0

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Common difference of A.P. must be 2b.

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Common difference of A.P . must be 2c.

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All above

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Solution

The correct options are
B All above
C $a = 0$
D Common difference of A.P . must be 2c.
$S_{n} = \frac{n}{2} [2 a^{'} + (n - 1) d] = a + b n + c n^{2}$
$\Rightarrow n a^{'} + \frac{n (n - 1) d}{2} = a + b n + c n^{2}$
$\Rightarrow (a^{'} - \frac{d}{2}) n + \frac{n^{2} d}{2} = a + b n + c n^{2}$
On comparing coefficient, we get
$a = 0, b = a^{'} - \frac{d}{2}, c = \frac{d}{2}$

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

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	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 $(D) = (2 b - 4)$

a n^{2} + b n

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