Answer:
Step-by-step explanation:
for each positive integer m, so that la - r2m-ll and la - r2m I are both less than
1 1
Ir2m - r2m-ll .:::: 42(m-l) Ir2 - rll = 24m - 3 ·
It follows that limn--->oo rn = limn--->oo rn-l = a. Taking limits in the recursion in
(d) leads to
1
a=I+--
l+a
which reduces to a 2 = 2. Since rn > 0 for each n, a > O.
Round the number like for you 270+350=620
Answer:
Your Answer is going to be C
A_n= a₁+(n-1)d
a₁ first term
n terms
d distance between each value
a_n= 12+(405-1)(5)=2032
