Question

Find the indicate term for each geometric series described.

Answer 1

\[Find: a _{1}\] \[a _{n}=-162\]\[s_{n}=-242\]\[r=3\]formular:\[s _{n}=\frac{a _{1(1-r^n)}}{1-r}\]

Answer 2

S = a( 1 - r^n)
---------
1 - r


r = 3
s = -242
l = -162

l = ar^n-1

s = a - a r * r^n-1
---------
-2

484 = a + 3 * 162

484 - 486 = a
-2 = a

Answer 3

how you get 484?

Answer 4

s*2

Answer 5

On the LHS there is -242 and RHS we get Denominator -2.... If I multiply -2 both sides...
I'l get 484 on LHS

Answer 6

I see .TY

Answer 7

can u come back explant me please!

Answer 8

Yea Sure :)

Answer 9

Tell me which part is confusing ...

Answer 10

s = a - a r * r^n-1
---------
-2


on the top how u come up:
a + 3 * 162 ?

Answer 11

I don know how u get n ?

Answer 12

a(1- r^n) right the numerator

Answer 13

yes, but how u get n ?

Answer 14

now u know gp it is
a, ar , ar^2.... ar^n-1
^
II
This is the last term
\[a _{n}\] in ur question this is the last term ... i took it as l

Answer 15

not I but L sry typo

Answer 16

now when we open a(1-r^n)

we get

a - ar^n

now i can write r^n as (r)* r^n-1....

Answer 17

this makes our eq...

a - (ar^n-1)(r)
now we know ar^n-1 is the last term ! don't we?

Answer 18

now the whole expression becomes( bfore this i was only doing it for numerator)


s = a - l r
-----
1-r

Answer 19

\[a(1-r^n)=a-ar^n\]

Answer 20

typo again sry.. its L not I
L for last term ......

Answer 21

your equation is right but the last term is a*r^n-1 not a*r^n

Answer 22

so when i try some cmn techniques .

ar^n-1(r) = ar^n

Answer 23

\[a \times r^{n-1}\]

Answer 24

tell me natal which part is confusin you ?

Answer 25

or wait a minute i'l post a solution attach file

Answer 26

I confuse , thequstion don't have n how to solve?

Answer 27

I will post one more similar to this question , can you help after done,TY

Answer 28

I be back