Question

calcule a soma de todos os termos da P.A.: (16, 20, 24,...,116)​

Answer 1

resolução!

r = a2 - a1

r = 20 - 16

r = 4

an = a1 + ( n - 1 ) r

116 = 16 + ( n - 1 ) 4

116 = 16 + 4n - 4

116 = 12 + 4n

116 - 12 = 4n

104 = 4n

n = 104/4

n = 26

Sn = ( a1 + an ) n / 2

Sn = ( 16 + 116 ) 26 / 2

Sn = 132 * 13

Sn = 1716

Answer 2

$Olá!!$

$\large\boxed{\boxed{{a_{n}=a_{1}+(n-1)r}}}}}$

Onde:

a1=16

an=116

r = a2—a1=20—16=4

n = ??

Logo teremos:

$116=16+(n-1)4$

$116=16+4n-4$

$116-16+4=4n$

$104=4n$

$\frac{104}{4}=n$

${\color{blue}{n=26}}$

Para soma temos que:

$\Leftrightarrow\:\:\large\boxed{\boxed{{S_{n}=\frac{a_{1}+a_{n}}{2}.n}}}}}$

$\Leftrightarrow\:S_{n}=\frac{16+116}{2}.26$

$\Leftrightarrow\:S_{n}=\frac{132.26}{2}$

$\Leftrightarrow\:S_{n}=\frac{3432}{2}$

${\color{blue}{\Leftrightarrow\:S_{n}=1716}}$

$\underline{Qualquer\:D\'uvida\:Comente}$