Question

Show that the ratio of sum of first n terms of GP to the sum of terms from (n+1)th term to (2n)th term is 1/r^n

Answer 1

\[S_n = a_1 \left(\dfrac{1 - r^n}{1 - r}\right)\]

Answer 2

\[a_{n + 1} = a_1 \cdot r\]

Answer 3

\[\LARGE S_n=\frac{a(r^n-1)}{r-1}\]
but what about 2nd one

Answer 4

how many terms in 2nd GP?

Answer 5

\[S= a_1 r\dfrac{}{}\]From \(n + 1\) to \(2n\), there are always \(n\) numbers.

Answer 6

why?

Answer 7

I mean\[S = a_1 r \left(\dfrac{1 - r^n}{1 - r}\right)\]And that "why?" has an easy answer. Dikhata hoon :-)

Answer 8

From \(1\) to \(n\), there are \(n\) numbers.