Question

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

(iv) If the sum of first n terms of this AP is 290, then the value of n is equal to:

[1 mark]

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

The correct option is D 10

$\text{Given,}$
$S_n=3n^2-n$
$\text{Also,}{S}_n=290$
$\therefore 3n^2-n=290$
$\Rightarrow 3n^2-n-290=0$
$\Rightarrow 3n^2-30n+29n-290=0$
$\Rightarrow 3n(n-10)+29(n-10)=0$
$\Rightarrow (n-10)(3n+29)=0$
$\Rightarrow (n-10)=0\text{and}(3n+29)=0$
$\Rightarrow n=10\text{and}n=\frac{-29}{3}\text{(rejected)}$
$\therefore n=10$