Sum of odd $n$ consecutive numbers $n^2$

The smallest prime number is $2$.

An even number is a number which has a factor of $2$.

Odd numbers are not divisible by $2$

A number is even if the digit in One's place is divisible by $2$.

Let ${S} _{n}=(2+4+6+.....+30)$. This is an A.P in which $a=2,d=2$ and $l=30$
Let the number of terms be $n$. Then,
$a+(n-1)d=30$
$\Rightarrow$ $2+(n-1)\times 2=30$
$\Rightarrow$ $n=15$
$\therefore$ ${S} _{n}=\cfrac{n}{2}(a+l)=\cfrac{15}{2}\times (2+30)=(15\times 16)=240$.

$-5,-4,-3,-2,-1,0,1,2,3,4,5$

$9\times 14=9\times 2\times 7$

Out of the following the integer which is positive and even is 4. Thus, option A is correct.

Any even integer is divisible by $2$.

If we add two even numbers the answer is always even.

$ Given\quad that\quad m,n,o,p\quad and\quad q\quad are\quad integers.\ \therefore \quad (n+o)\quad is\quad an\quad integer\ \quad \quad (p-q)\quad is\quad an\quad integer\ \therefore \quad m(n+o)(p-q)\quad is\quad a\quad product\quad of\quad 3\quad integers.\ The\quad product\quad to\quad be\quad even\quad one\quad of\quad the\quad factors\quad to\quad be\quad even.\ either\quad m\quad be\quad even\longrightarrow option\quad C\ or\quad (n+o)\quad be\quad even\longrightarrow no\quad option\ or\quad (p-q)\quad be\quad even\longrightarrow no\quad option.\ \therefore \quad option\quad C\quad is\quad the\quad answer. $

If we add two odd numbers the sum is always even.

Suppose $k$ is even

The number which ends with $2,4,6,8,0$ are divisible by $2$.

The numbers $1,3,5,7,9$ etc are called odd number which is not divisible by $2$ and any number which ends which these numbers are also called odd number.