Question

encontre o número e a soma dos termos das progressões aritméticas:

1- (4,7,10,...,94)
2- (1896,1900,1904,...,2016)
3- (3,9,15,...,333)
4- (7,3,-1,...,-269)
5- (-4,-10,-16,...,-2998)

Answer 1

1
a1 = 4
a2 = 7
r = 7 - 4 = 3 **
an =94
Sn = ( 4 + 94).n/2
Sn = 98 . n/2
Sn = 98n/2
Sn = 49n ***
an = 4 + ( n - 1). 3
94 =  4 + 3n - 3
94 = 1 + 3n
94 - 1 = 3n
3n =93
n = 93/3 = 31 ***
Sn = 49 * 31 
S31 = 1519 ****

2
a1 = 1896
a2 = 1900
r = 1900 - 1896 = 4 ***
an = 2016
2016 = 1896+ ( n - 1)4
2016 = 1896 + 4n - 4
2016 = 1892 + 4n
2016 - 1892 = 4n
4n = 124
n = 124/4 = 31 ***
S31 = ( 1896 + 2016).15.5
S31 = 60636 ****
3
a1 = 3
a2 = 9
an = 333
r = 9 - 3 = 6 **
333 = 3 + ( n - 1)6
333 = 3 + 6n - 6
333 = 6n - 3
333 + 3 = 6n
6n = 336
n = 336/6 =56 ***
S56 = ( 3 + 333).28
S56 = 336 * 28
S56 = 9408 ****
4
a1 = 7
a2 = 3
an = - 269
r = 3 - 7 = - 4 ***
- 269 =  7 + ( n - 1)( - 4)
-269 =  7 + ( -4n + 4)
- 269 = 11 - 4n
-269 - 11 =  - 4n
- 280 = - 4n
4n = 280
n = 280/4 = 70 ****

S70 = ( 7 - 269).35
S70 = - 262 . 35
S70 = - 9170 ****

5
a1 = -4
a2 = - 10
r = -10 - ( -4) = -10 + 4 = - 6 ****

an = -2998
- 2998 =  - 4 + ( n - 1)(-6)
- 2998 = -4 - 6n + 6
-2998 = 2 - 6n
-2998 - 2 = - 6n
6n = 3000
n = 3000/6 = 500 
S500 = ( -4 - 2998). 250
S500 = - 3002 . 250
S500 = 750500 **