Question 1177555: Please I will Cashapp 10$ If anyone can solve it it's due today I really need help.
Pick DIFFERENT values that meet each description.
m = ____ (m is an integer and m<-10) d = ____ (d is an integer and -10 < d < -1)
n = ____ (n is an integer and n > 20) r = ____ (r is a rational # and 0 < r < 1)
1) Create an arithmetic sequence with a common difference of d, and the third term in the sequence is m. Show the first 5 terms of the sequence.
2) Write the explicit rule for the sequence that you created. Use your rule to confirm the 5th term of the sequence.
3) Write the recursive rule for your arithmetic sequence.
4) Use the formula for arithmetic series, and find the sum of the first 10 terms of the sequence that you created.
Part 2:
1) Using your chosen value of n, create a geometric sequence with a common ratio of r, and the first term of the sequence is n. Show the first 5 terms of the sequence.
2) Write the explicit rule for the sequence that you created. Use your rule to find the 10th term of the sequence.
3) Write the recursive rule for your geometric sequence.
4) Write a summation notation that depicts the sum of the first 25 terms of the geometric series. Find the sum. (Round to 5 decimal places if necessary.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Arithmetic:
m is -25 d=-5
-15,-20,-25,-30,-35
-
this is an=a1+(n-1)d=-15+(n-1)(-5)
series is 15,-20,-25,-30,-35
a5=-15+4(-5)=-35
-
an=a(n-1)-5
-
Sn=(n/2)(2a1+(n-1)d)
S10=5(-30+9(-5))=-375
series of -15,-20,-25,-30,-35,-40,-45,-50,-55,-60
--
n=625 and r=0.2 or 1/5
625,125,25,5,1
rule is a(n)=a(n-1)*(1/5)
an=ar^(n-1) so a10=(.2^9)=0.00032, this is also, continuing from above(1/5+(1/25)+(1/125)+(1/625)+(1/3125)), where 1/3125 is the 10th term of the sequence. Note: 1/3125 is 0.00032.
-
summation is S25=(a (1-r^25)/(1-r))
625(1-.2^25/.8=781.25
This is the limit of the series, which is a/(1-r)=625/.8=781.25
|
|
|
