OpenStudy (anonymous):

Last ones I need help with!

3 years ago
OpenStudy (anonymous):

3 years ago
hartnn (hartnn):

so the common difference =d= +2 a1 = January clients = 5 use the formula \(\Large a_n = a_1 +(n-1)d\)

3 years ago
hartnn (hartnn):

december means n = 12

3 years ago
OpenStudy (anonymous):

Alright so a_n=a_1(n-1)d a_n=5+(n-1)2 a_n=5+(12-1)2 a_n=5+11(2) a_n=5+22 a_n=27

3 years ago
hartnn (hartnn):

correct, 27 clients in december

3 years ago
hartnn (hartnn):

to get the total number of clients in 12 months, we need to add all the clients she added each month! we could use this sum formula \(\Large S_n = (n/2) (2a+(n-1)d)\) n = 12

3 years ago
hartnn (hartnn):

that \(a\) is actually \(a_1\)

3 years ago
OpenStudy (anonymous):

S_n=(n/2)(2a_1+(n-1)d) S_n=(12/2)(2(5)+(12-1)2) S_n=(6)(10+(11)2) S_n=(6)(10 + 22) S_n=(6)(32) S_n=192

3 years ago
hartnn (hartnn):

correct :) ask if any doubts...

3 years ago
OpenStudy (anonymous):

Hahaa. Okay great so I don't know how to do the sigma ones at all.

3 years ago
hartnn (hartnn):

\(\Large \sum = sum\) \(\sum \limits_{n=1}^6\) means sum from 1st term to 6th term

3 years ago
hartnn (hartnn):

one easier approach to do both 2 and 3, (just because the upper limit is small) is to keep on plugging in numbers for n plug in n =1, then n=2, then n=3....n=6 and add all the answers

3 years ago
hartnn (hartnn):

for 3. start with n = 0, go till n =5 add all the answers

3 years ago
OpenStudy (anonymous):

So for 2 n^2+1 1^2+1=1+1=2 2^2+1=4+1=5 3^2+1=9+1=10 4^2+1=16+1=17 5^2+1=25+1=26 6^2+1=36+1=37 2+5+10+17+26+37=97

3 years ago
hartnn (hartnn):

too lazy to check, i assume thats correct :P

3 years ago
hartnn (hartnn):

yup, 97 is correct

3 years ago
OpenStudy (anonymous):

Lol okay For 3 16(1/4)^n 16(1/4)^0=16*1=16 16(1/4)^1=16*.25=4 16(1/4)^2=16*.0625=1 16(1/4)^3=16*.015625=.25 16(1/4)^4=16*.00390625=.0625 16(1/4)^5=16*9.765625E-4=.015625 16+4+1+.25+.0625+.015625=21.328125

3 years ago
hartnn (hartnn):

correct! :)

3 years ago
OpenStudy (anonymous):

Ok so how do I determine the last one.

3 years ago
hartnn (hartnn):

can you find the common ratio between the terms of the last sequence ?

3 years ago
hartnn (hartnn):

if magnitude of common ratio |r| > 1 , then the series does not converge at all and the sum would be infinite

3 years ago
OpenStudy (anonymous):

So to find r Do I do it like this? 1/4 to 1/2 or is it 1/4 to -1/2?

3 years ago
hartnn (hartnn):

r = 2nd term / 1st term = 3rd term/ 2nd term = ...

3 years ago
hartnn (hartnn):

-1/2 divided by 1/4 or simply -2 divided by 1

3 years ago
OpenStudy (anonymous):

-1/2 divided by 1/4 = -2 -2 divided by 1= -2 |r|>1 |-2|>1 2>1

3 years ago
hartnn (hartnn):

correct, so the sum does not even exist, as the sequence is divergent

3 years ago
OpenStudy (anonymous):

Alright thank you!

3 years ago
hartnn (hartnn):

welcome ^_^

3 years ago
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