# Find first term and ratio of endless geometric sequence

An endless geometric sequence is one where its ratio: $$r \neq 0, -1 \lt r \lt 1$$

The sum of an endless geometric sequence is followed by the following formula: $$S_n = \frac{a_1}{1 - r}$$

From the endless geometric sequence $a_1,a_2,a_3,...$ that the ratio is $-1 \lt r \lt 1$, they created a new endless geometric sequence $a_1 + a_2 + a_3, a_3 + a_4 + a_5, a_5+a_6+a_7,...$ The sum of the new sequence is 1.9 times bigger than the sum of the original sequence. Find the ratio of the original sequence.

I don't know even how to begin so I can't show any tries.

Our new geometric sequence is $$(a+ar+ar^2)+(ar^2+ar^3+ar^4)+(ar^4+ar^5+ar^6)+(ar^6+ar^7+ar^8)+\cdots.$$ Add up. We get the old sum, plus an additional $ar^2+ar^4+ar^6+\cdots$.
From the information given, we can deduce that $\dfrac{ar^2}{1-r^2}=(0.9)\dfrac{a}{1-r}$. The rest should be straightforward.
Hint $$a_{2n+1}+a_{2n+2}+a_{2n+3}=a_1r^{2n}+a_2r^{2n}+a_3r^{2n}=(a_1+a_2+a_3)(r^2)^n$$
• Can you explain how did you get to $a_{2n+1} + a_{2n+2} + ...$? – Pichi Wuana Jan 6 '16 at 18:48
• He applied the definition of the geometric series. $a_{2n+1}$ is $2n$ steps along, so $a_{2n+1}=a_1r^{2n}$. The others are similar. – Ross Millikan Jan 6 '16 at 18:50