Question

The sum of n terms of an arithmetic progression (AP) is $3n^2-n.$

(i) The first term and the common difference of the given AP are:


[1 mark]

• A. 2, -6
• B. 4, 2
• C. 2, 6
• D. 2, 4

Answer 1

The correct option is C 2, 6

Let a be the first term of the AP, and d be the common difference.
Given,
$S_n=3n^2-n$
$\Rightarrow \frac{n}{2}[2a+(n-1\left)d\right]$$=3n^2-n\{\because S_n=\frac{n}{2}[2a+(n-1)d\left]\right\}$
$\Rightarrow n[2a+(n-1\left)d\right]=6n^2-2n$
$\Rightarrow 2an+n(nd-d)=6n^2-2n$
$\Rightarrow n^{2}d+(2a-d)n=6n^2-2n$
On comparing the coefficient of $n^2$, we get,
d = 6
On comparing the coefficient of n, we get,
$(2a-d)=-2$
$\Rightarrow 2a=-2+d=-2+6\{\because d=6\}$
$\Rightarrow 2a=4$
$\Rightarrow a=2$