Geometric sequence

**megs_world** · January 25th 2009, 05:54 PM

In a geometric sequence, t1 + t2 + t3 = 21 and t4 + t5 + t6 =168. Find the sequence.

how do u use the formula Sn = a(r^n -1)/r -1) to find the sequence?

**Soroban** · January 25th 2009, 07:43 PM

Hellp, megs_world!

In a geometric sequence: . $t_1 + t_2 + t_3 \:=\: 21\,\text{ and }\,t_4 + t_5 + t_6 \:=\:168$

Find the sequence.

how do u use the formula: $S_n \:=\: a\frac{r^n -1}{r -1}$ to find the sequence?

We know the sum of the first 3 terms: . $S_3 \:=\:21$
. . Hence: . $a\frac{r^3-1}{r-1} \:=\: 21$ .[1]

We know the sum of the first 6 terms: . $(t_1+t_2+t_3) + (t_4+t_5+t_6) \:=\:21 + 168 \:=\:189$
. . Hence: . $a\frac{r^6-1}{r-1} \:=\:189$ .[2]

Divide [2] by [1]: . $\frac{a\dfrac{r^6-1}{r-1}}{a\dfrac{r^3-1}{r-1}} \:=\:\frac{189}{21} \quad\Rightarrow\quad \frac{r^6-1}{r^3-1} \:=\:9$

Factor and reduce: . $\frac{(r^3-1)(r^3+1)}{r^3-1} \:=\:9 \quad\Rightarrow\quad r^3 + 1 \:=\:9$

. . Then: . $r^3 \:=\:8 \quad\Rightarrow\quad\boxed{ r \:=\:2}$

Substitute into [1]: . $a\frac{2^3-1}{2-1} \:=\:21 \quad\Rightarrow\quad7a \:=\:21 \quad\Rightarrow\quad\boxed{ a \:=\:3}$

Therefore, the sequence is: . $t_n \:=\:3\cdot2^{n-1} \quad\Rightarrow\quad 3, 6, 12, 24, 48, 96\hdots$

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