Verified answer

1) an= a1 .q^n-1

8192= 4 . 2^(n-1)

2048= 2^(n-1)

2^11= 2^ n-1

11= n-1

n= 12

2) 501 e 999

999= 501 + (n-1).3

498= 3n-3

501= 3n

n= 167

8192=4*2^n-1

8192/4=2^n-1

2048=2^n-1

2^11=2^n-1

n-1=11

n=11+1

n=12

12 termos

1-Quantos termos tem a PG (4,8,16...8192)

an = 8.192

a1 = 4

n = ?

q = a2 : a1 = 8 : 4 = 2

an = a1*q^n-1

8.192 = 4 * 2^n-1

8.192 / 4 = 2^n-1

2.048 = 2^n-1

2¹¹ = 2^n-1

n - 1 = 11

n = 11 + 1

n = 12

2-Quais são os múltiplos de 3 compreendidos entre 500 e 1000?

an = 999

a1 =501

r = 3

n = ?

an = a1 + (n-1)r

999 = 501 +(n-1)3

999 = 501 + 3n - 3

999 = 498 + 3n

3n = 999 - 498

3n = 501

n = 167