Question

In a GP with n terms, the first term is 1, the common ratio is r and the sum is s, where r and s are not zero. If each term of the GP is replaced by its reciprocal, then the sum becomes?

Answer 1

@ganeshie8

Answer 2

@perl

Answer 3

Would it not diverge then?

Answer 4

example 1/2+1/4+1/8....would become 2+4+8....?

Answer 5

im not sure about that, just basic AP, GP

Answer 6

I am a little out of my realm because I only took basic calc series and never studied progressions in general, but it seems from the example I showed, if I understand you correctly, that we cant guarantee it converges.

Answer 7

ok

Answer 8

how should i start

Answer 9

\[\large 1+r+r^2+\cdots + r^{n-1} = s\]

Answer 10

replace the left side with partial sum formula

Answer 11

\[\large \dfrac{1-r^n}{1-r} = s\]

Answer 12

what about the reciprocals @ganeshie8

Answer 13

if you replace each term by their reciprocal, the series becomes:

\[\large 1+\frac{1}{r}+\frac{1}{r^2}+\cdots +\frac{1}{ r^{n-1}} \]

Notice that the common ratio is 1/r now

Answer 14

i still dont know how to get the relation

Answer 15

partial sum would be
\[\large \dfrac{1 - \frac{1}{r^n}}{1-\frac{1}{r}}\]

rearragne a bit and try to express it in terms of \(s\)

Answer 16

is it r^n/s

Answer 17

nope, try again

Answer 18

not sure

Answer 19

\[\large \dfrac{1 - \frac{1}{r^n}}{1-\frac{1}{r}} \]

\[\large \dfrac{(r^n - 1)/r^n}{(r-1)/r} \]

\[\large \dfrac{1-r^n}{1-r} \cdot \dfrac{r}{r^n} \]

\[\large s \cdot \dfrac{1}{r^{n-1}} \]

Answer 20

ohh right

Answer 21

thanks

Answer 22

so its s/r^n-1

Answer 23

Yep!