Answer:
ob is a ray between OC and OA... may it help you
C. 7/8 is larger than 1/5
Yes it is possible for a geometric sequence to not outgrow an arithmetic one, but only if the common ratio r is restricted by this inequality: 0 < r < 1
Consider the arithmetic sequence an = 9 + 2(n-1). We start at 9 and increment (or increase) by 2 each time. This goes on forever to generate the successive terms.
In the geometric sequence an = 4*(0.5)^(n-1), we start at 4 and multiply each term by 0.5, so the next term would be 2, then after that would be 1, etc. This sequence steadily gets closer to 0 but never actually gets there. We can say that this is a strictly decreasing sequence.
If your teacher insists that the geometric sequence must be strictly increasing, then at some point the geometric sequence will overtake the arithmetic one. This is due to the nature that exponential growth functions grow faster compared to linear functions with positive slope.
Step 1
calculate the value for i=1
=(4*(1/2)<span> ^(1-1))---> 4
step 2
</span>calculate the value for i=2
=(4*(1/2) ^(2-1))---> 2
step 3
calculate the value for i=
=(4*(1/2) ^(3-1))---> 4*(1/4)---->1
step 4
find the sum
(4+2+1)=7
