Geometric Sequence (Only 4 terms and their sums are given)

[CrimsonKnight]

Homework Statement

"In a geometric sequence, the sum of t₇ and t₈ is 5832, the sum of t₂ and t₃ is 24. Find the common ratio and first term."

Homework Equations

d = t₈/t₇ or t₃/t₂
t_n = a * r^n-1

The Attempt at a Solution

So I thought of developing a system of equations then solving by elimination. The result that I got is r = 3.8 but I have a gut feeling that it's wrong because when I plug it into the equation, it gives me unbelievably huge numbers and the end result would be negative and I believe I did something wrong when trying to find a.

Explaining the reasoning behind the solution would be appreciated :)

 

[NasuSama]

Did you show the work here? I asked this question because I want to see how you approached this problem.

Other than that, the answer is not $3.8$. I recommend you start this problem by indicating all available info you have.

 

[CrimsonKnight]

Did you show the work here? I asked this question because I want to see how you approached this problem.

Other than that, the answer is not $3.8$. I recommend you start this problem by indicating all available info you have.

I'm also asking help on how to approach this problem. Hope someone can enlighten us on how to solve this problem.

 

[NasuSama]

We can help you, but we can't answer the question for you since it's important for students to work out the problem by themselves.

Based on the problem, we know that since the sum of $t_7$ and $t_8$ is $5832$ and the sum of $t_3$ and $t_2$ is $24$, we have

$t_2 + t_3 = 24$
$t_7 + t_8 = 5832$

Since $t_n$ is a geometric sequence, we obtain

$ar^{2 - 1} + ar^{3 - 1} = 24$
$ar^{7 - 1} + ar^{8 - 1} = 5832$

$ar + ar^{2} = 24$
$ar^{6} + ar^{7} = 5832$

Let's see if you can solve for $r$.

 

[CrimsonKnight]

We can help you, but we can't answer the question for you since it's important for students to work out the problem by themselves.

Based on the problem, we know that since the sum of $t_7$ and $t_8$ is $5832$ and the sum of $t_3$ and $t_2$ is $24$, we have

$t_2 + t_3 = 24$
$t_7 + t_8 = 5832$

Since $t_n$ is a geometric sequence, we obtain

$ar^{2 - 1} + ar^{3 - 1} = 24$
$ar^{7 - 1} + ar^{8 - 1} = 5832$

$ar + ar^{2} = 24$
$ar^{6} + ar^{7} = 5832$

Let's see if you can solve for $r$.

Using the elimination process for solving systems of equations:
$ar^{6} + ar^{7} = 5832$
$ar + ar^{2} = 24$

$r^{5} + r^{5} = 5808$
$2r^{5} = 5808$
$r^{5} = 243$
$r = \sqrt[5]{243}$
$r = 3$

Substituting $r = 3$ into the previous equation:
$ar + ar^{2} = 24$
$a = 24/ (r + r^{2})$
$a = 24/ (3 + 3^{2})$
$a = 24/(3+9)$
$a = 2$

Now I understand; I made the problem of subtracting 24 to 5832 when I should've divided it instead.
Thank you very much :)

 

[NasuSama]

Very well done!