Question

The least value of n such that $1+3+5+\mathrm{7...}n$ terms $\ge 500$ is?

• A. $18$
• B. $19$
• C. $22$
• D. $23$

Answer 1

The correct option is D $23$

Given:

$1+3+5+7\cdots +n$

Solution:

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

according to question,

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)\ge 500$

$\Rightarrow \frac{n}{2}\left(2\right(1)+(n-1\left)\right(2\left)\right)\ge 500$

$\Rightarrow \frac{n}{2}(2+(n-1\left)\right(2\left)\right)\ge 500$

$\Rightarrow n(1+(n-1\left)\right)\ge 500$

$\Rightarrow n\left(n\right)\ge 500$

$\Rightarrow n^2\ge 500$

Now we have 18, 19, 22 and 24

$\therefore (18{)}^2=324$

$(19{)}^2=361$

$(22{)}^2=484$

$(23{)}^2=529$

We get ,

$(23{)}^2>500$

$\therefore n=23$