A=84 and r=1/5
Since r, the common ratio squared is less than one, the sum will converge to a limit. Rule: if r^2<1 infinite series converges, otherwise it diverges.
Since the sum of any geometric sequence is:
s(n)=a(1-r^n)/(1-r)
And if r^2<1, (1-r^n) becomes 1-0=1 as n approaches infinity.
So whenever r^2<1 the sum of the infinite series is just:
s(n)=a/(1-r)
Since a=84 and r=1/5 this infinite series has a sum of:
s(n)=84/(1-1/5)
s(n)=84/(4/5)
s(n)=5(84)/4
s(n)=105
Answer: 469762048
Step-by-step explanation:
1: 7*4^0=
7*1=7
2:7*4^1=
7*4=28
3: 7*4^(3-1)=
7*4^2=
7*16=112
14: 7*4^(14-1)=
7*4^13=469762048
The greatest common The greatest common factor in my words: The greatest of the common factors of two or more numbers.
Sosksl 35is40 is 40 times 0 is 3
