Question

PA - 54 pontos

2) Calcule a soma dos 24 primeiros de cada PA
a) (-57, -27, 3, ...)
b) (7, 7, 7, ...)
c) (2/3, 8/3, 14/3, ...)
d) (-1/2, -1/4, 0, ...)
3) Quantos termos da PA (3,19,35,...) devem ser somados para que a soma Sn=472?


Necessito disto urgentemente!

Answer 1

Juliana,
As 4 de 2) são da mesma natureza. Tem o  mesmo processo de solução
Vou resolver c)
Com essa base, as outras levam poucos minutos

c)
         S24 = 24/2(2/3 + a24)
         2/3, 8/3, 14/3
           r = 6/3 = 2       (r = 8/3 - 2/3 = 14/3 - 8/3 = 6/6
           a24 = 2/3 + (24 - 1).(2)
                   = 2/3 + 46
                   = 2/3 +138/3
           a24 = 140/3
                                         S24 = 24/2(2/3 + 140/3)
                                                 = 12(142/3)
                                                 = 1704/3
                                                                     S24 = 568
 

Answer 2

2) Calcule a soma dos 24 primeiros de cada PA:

1° calcular an de cada PA.

Fórmulas:

$a_n=a_1+(n-1)\times r \\\\\\ S_n= \dfrac{(a_1+a_n)\times n}{2}$



$a)~~(-57, -27, 3, ...)\\\\a_{24}=-57+(24-1)\times 30\to~~ a_{24}=-57+23\times 30\to\\\\ a_{24}=-57+690\to \boxed{a_{24}=633}\\\\\\\\ S_{24}= \dfrac{(-57+633)\times24}{2} \to~~ S_{24}= \dfrac{576\times24}{2} \to~~S_{24}= \dfrac{13824}{2} \to~~ \\\\ \large\boxed{\boxed{S_{24}= 6912}}$





$b)~~ (7, 7, 7, ...)\\\\a_{24}=7+(24-1)\times 0\to~~ a_{24}=7+23\times 0\to\\\\ a_{24}=7+0\to \boxed{a_{24}=7}\\\\\\\\ S_{24}= \dfrac{(7+7)\times24}{2} \to~~ S_{24}= \dfrac{14\times24}{2} \to~~ S_{24}= \dfrac{336}{2} \to~~ \\\\\large\boxed{\boxed{S_{24}= 168}}$





$c)~~\left( \dfrac{2}{3} , \dfrac{8}{3} , \dfrac{14}{3} , ...\right)\\\\a_{24}=\dfrac{2}{3}+(24-1)\times 2\to~~ a_{24}=\dfrac{2}{3}+23\times 2\to\\\\ a_{24}=\dfrac{2}{3}+46\to~~ a_{24}=\dfrac{2+138}{3}\to~~ \boxed{a_{24}=\dfrac{140}{3}}\\\\\\\\ S_{24}= \dfrac{\left(\dfrac{2}{3}+\dfrac{140}{3}\right)\times24}{2} \to~~ S_{24}= \dfrac{\dfrac{142}{3}\times24}{2} \to~~ S_{24}= \dfrac{\dfrac{3408}{3}}{2} \to~~ \\\\ S_{24}=\dfrac{1136}{2} \to~~\large\boxed{\boxed{S_{24}= 568}}$





$d)~~\left(- \dfrac{1}{2} ,- \dfrac{1}{4} , 0, ...\right) \\\\a_{24}=- \dfrac{1}{2}+(24-1)\times \dfrac{1}{4} \to~~ a_{24}=- \dfrac{1}{2}+23\times \dfrac{1}{4} \to\\\\ a_{24}=- \dfrac{1}{2}+ \dfrac{23}{4} \to~~ a_{24}=\dfrac{-2+23}{4}\to~~ \boxed{a_{24}=\dfrac{21}{4}}\\\\\\\\ S_{24}= \dfrac{\left(- \dfrac{1}{2}+\dfrac{21}{4}\right)\times24}{2} \to~~ S_{24}= \dfrac{\dfrac{19}{4}\times24}{2} \to~~ S_{24}= \dfrac{\dfrac{456}{3}}{2} \to~~ \\\\ S_{24}=\dfrac{152}{2} \to~~\large\boxed{\boxed{S_{24}= 76}}$







3) Quantos termos da PA (3,19,35,...) devem ser somados para que a soma Sn=472?

$a_1=3\\r=16\\s_n=472\\ a_n=a_1+(n-1)\times r\\n=?$


$472= \dfrac{(3+a_n)\times n}{2} \to\\\\\\ 472= \dfrac{(3+a_1+(n-1)\times r)\times n}{2} \to\\\\\\ 472= \dfrac{(3+3+(n-1)\times 16)\times n}{2} \to\\\\\\ 472= \dfrac{(6+16n-16)\times n}{2} \to\\\\\\ 472= \dfrac{(16n-10)\times n}{2} \to\\\\\\ 472= \dfrac{16n^2-10n}{2} \to~~ multiplique~~cruzado:$



$944= 16n^2-10n \to~~ 16n^2-10n -944=0\\\\a=16;~b=-10;~c=-944\\\\ \Delta=b^2-4ac\to \Delta=(-10)^2-4.16.(-944)\to \Delta=100+60416\to\\ \Delta=60516\\\\ x= \dfrac{-b\pm \sqrt{\Delta} }{2a} \to~~ x= \dfrac{-(-10)\pm \sqrt{60516} }{2.16} \to~~ x= \dfrac{10\pm 246 }{32} \to\\\\\\ x'= \dfrac{10+ 246 }{32} \to~~ x'= \dfrac{ 256 }{32} \to~~ \boxed{x'=8}\\\\ x''= \dfrac{10- 246 }{32} \to~~ x''= \dfrac{ -236 }{32} \to~~ \boxed{x'=- \dfrac{59}{8}}$



Desprezamos a raiz negativa, então teremos 8 termos para que a soma dessa PA seja igual a 472.