Question

The value of x for which the sum $1+4+7+10+....+x=590$ is .

• A. 54
• B. 48
• C. 58
• D. 52

Answer 1

The correct option is C 58

$1+4+7+10+....+x=590$
This is an AP with first term 1,common difference 3. Let x be the nth term of the AP. Therefore, sum of n terms of AP is 590.
$S_n=590\Rightarrow \frac{n}{2}\left(2a+\left(n-1\right)d\right]=590\Rightarrow \frac{n}{2}\left(2\times 1+\left(n-1\right)\times 3\right]=590\Rightarrow \frac{n}{2}\left(2+3n-3\right]=590\Rightarrow 3n^2-n-1180=0\Rightarrow 3n^2-60n+59n-1180=0\Rightarrow \left(3n+59\right)\left(n-20\right)=0\Rightarrow n=-\frac{59}{3}\left(\mathrm{Not}\mathrm{Possible}\right)\mathrm{or}n=20\therefore x\mathrm{is}20\mathrm{th}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{AP}.a_{20}=x\Rightarrow 1+(20-1)3=x\Rightarrow x=58$