Question

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Answer 1

@ganeshie8 @satellite73 help

Answer 2

first term = \(\large a_1 = 740\)

common ratio = \(\large r = \dfrac{1}{6}\)

Answer 3

Can you write the general term for this geometric sequence ?

Answer 4

http://education-portal.com/cimages/multimages/16/Geometric_Sequence_Rule.png

Answer 5

use that^

Answer 6

Ok, so the general term is an = a1r^n-1

Answer 7

yes,plugin \(a_1\) and \(r\) values

Answer 8

general term = \(\large a_1r^{n-1} = 740 \left(\dfrac{1}{6}\right)^{n-1}\)

Answer 9

an = 740 (1/6)^n - 1
?

Answer 10

Yes !

Answer 11

knw how to represent the infinite sum in sigma notation ?

Answer 12

infinite sum = \(\large \sum \limits_{n=1}^{\infty} 740 \left(\dfrac{1}{6}\right)^{n-1}\)

Answer 13

thats the sum in sigma notation

Answer 14

yes, I just was writing that. Is the sum divergent ?

Answer 15

nope

Answer 16

the sum is convergent because the common ratio is < 1

Answer 17

use the infinite geometric series formula to calculate the sum

Answer 18

https://cdn.tutsplus.com/active/uploads/legacy/tuts/122_physics/tutorial/sum-geometric-series.jpg

Answer 19

use that formula^

Answer 20

Ok....

I'm getting 891.56 which can't be right
One of the options is 888 , maybe this is it since it's closest?

Answer 21

@ganeshie8 ?

Answer 22

check ur calculation again :
http://www.wolframalpha.com/input/?i=740%2F%281-1%2F6%29