Question

Find the mean of first n even positive integers.

Answer 1

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\text{first n even positive integersare,}$

$\sf\dashrightarrow 2+4+6+.......n_{th} term.$

$\sf\dashrightarrow a= 2$

$\sf\dashrightarrow d= t_2-t_1 \\ \implies d= 4-2\\ \implies d=2$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow\text{ mean of first n even positive integers.}$

✯FORMULA IN USE,

$\large{\boxed{\bf{ \star\:\: S_n= \dfrac{n}{2} \bigg( 2a+(n-1)d \bigg)\:\: \star}}}$

$\large\underline\bold{SOLUTION,}$

$\sf\dashrightarrow\text{ putting the values in the formula and solving it,}$

$\sf\dashrightarrow S_n= \dfrac{n}{2} \times \bigg( 2(2)+(n-1)(2) \bigg)$

$\sf\implies S_n= \dfrac{n}{2} \times \bigg(4+(2n-2) \bigg)$

$\sf\implies S_n= \dfrac{n}{2} \times \bigg(2+2n\bigg)$

$\sf\implies S_n= \dfrac{n}{2} \times 2\big(1+n\big)$

$\sf\implies S_n= \dfrac{n}{\cancel{2}} \times \cancel{2}\:\:\big(1+n\big)$

$\sf\implies S_n= n(n+1)$

$\sf\therefore\text{ the n th term is, n(n+1).}$

NOW,

$\large\therefore mean= \dfrac{S_n}{n}$

$\sf\implies \dfrac{ n(n+1)}{n}$

$\sf\implies \dfrac{\cancel{n}(n+1)}{\cancel{n}}$

$\sf\implies n+1$

$\large{\boxed{\bf{ \star\:\: mean= n+1\:\: \star}}}$

$\large\underline\bold{MEAN \:OF\: FIRST \:n\: EVEN\: POSITIVE\: INTEGERS\:IS\:n(n+1)}$

___________

Answer 2

Answer:

Answer N+1 hai zyada explanation ki jaroorat nhi pura panna bhar de rahe bande log