Question

the limiting sum of a geometric series is 5 and the 2nd term is 6/5. find the first term and the common ratio.

Answer 1

sorry bt can u say again wht's the role of 5

Answer 2

its the limiting sum

Answer 3

so its that

Answer 4

thats the limiting sum formula

Answer 5

i'm doing geometric series bt haven't yet came across limiting sum!

Answer 6

you should come across it soon

Answer 7

ahan

Answer 8

Now if you're given the second term,which is 6/5, hw would you use this to find the first term in the sequence? :)

Answer 9

not sure

Answer 10

what the name of this formula ?

Answer 11

geometric series.

Answer 12

like first term is \(a_1\), 2nd term, you just multiply by r, 2nd term = \(a_1r\), 3rd term, you again multiply by r, 3rd term = \(a_1r^2\)
terms ----> \(a_1,a_1r,a_1r^2,a_1r^3....\)
so, general term = \(a_n=a_1r^{n-1}\)
here you have
ar = 6/5
and a/(1-r) =5
solv simultaneously...

Answer 13

hmm...

Answer 14

a= 6/(5r)

Answer 15

ar = 6/5 therefore a = 6/5r lol..... -_-

\[\huge \frac{\frac{6}{5r}}{1-r}=5\]

Answer 16

yes^

Answer 17

@yasmin95 could u solve now ?

Answer 18

yes thanks :)

Answer 19

welcome ^_^

Answer 20

\[\large \frac{6}{5r}= 5-5r\]\[\large 5r + \frac{6}{5r}=5\] Im too l azy.

Answer 21

quadratic equation...i guess yasmin already solved it :)

Answer 22

I don't even know.....how to.... :| havent bothered, lol.

Answer 23

@yasmin95, how did you solve thiss