1. ## Sum to Infinity

Hi again guys!

Once again I need help with what should be relatively simple, because once again my maths textbook refuses to explain things properly

Ok so its about the Sum to Infinity of a Geometric Series:

In a certain geometric sequence the first term exceeds the second term by 32 and the sum of the second and third terms is 48. Find the infinite sum of the sequence.

I thought I had the answer, but turns out I was way off-I'll give you what I got and I would appreciate it if someone could please tell me where I went wrong. I got: a= 192/7, r=-1/6

Because a=32/1-r and a= 48/r-r^2. Since a=a, 32/1-r = 48/r-r^2

I solved for r by using the Quadractic formula.

Thankssss!!

2. Hello,

Originally Posted by Stevo_Evo_22
Hi again guys!

Once again I need help with what should be relatively simple, because once again my maths textbook refuses to explain things properly

Ok so its about the Sum to Infinity of a Geometric Series:

In a certain geometric sequence the first term exceeds the second term by 32 and the sum of the second and third terms is 48. Find the infinite sum of the sequence.

I thought I had the answer, but turns out I was way off-I'll give you what I got and I would appreciate it if someone could please tell me where I went wrong. I got: a= 192/7, r=-1/6

Because a=32/1-r and a= 48/r-r^2. Since a=a, 32/1-r = 48/r-r^2

I solved for r by using the Quadractic formula.

Thankssss!!
I'm not sure about what you did....

Let $a_1$ be the first term of the geometric series, and r the geometric progression.

$a_1=a_2+32$
But we also know that $a_2=r \cdot a_1$

$\implies \boxed{a_2=r \cdot a_2+32 \cdot r}$

We know that $a_2+a_3=48$, but we also know that $a_3=r \cdot a_2$

$\implies \boxed{a_2+r \cdot a_2=48}$

$\implies a_2 \cdot (1+r)=48 \implies a_2=\frac{48}{1+r}$

Substituting this into the first boxed equation, we get :

$\frac{48}{1+r}=r \cdot \frac{48}{1+r}+32 \cdot r$

Multiply by 1+r and solve

The infinite sum is $S=a_1 \cdot \frac{1}{1-r}$ if $|r|<1$

3. Whoaa, Ok now I'm very confused. How do you get "a(lower 2) = r x a(lower 2) +32r"????

Yes, I do need it in simple terms

4. Originally Posted by Moo
$a_1=a_2+32$
But we also know that $a_2=r \cdot a_1$

$\implies \boxed{a_2=r \cdot a_2+32 \cdot r}$

I substitute a_1 of the first equation in the second

5. Originally Posted by Moo
[snip]
Let $a_1$ be the first term of the geometric series, and r the common ratio for the geometric progression.

$a_1=a_2+32$
But we also know that $a_2=r \cdot a_1$

$\implies \boxed{{\color{red}a_1 =r \cdot a_1 +32}}$ .... (1)

We know that $a_2 + a_3=48$, but we also know that $a_3 = r \cdot a_2$

$\implies a_2 + r \cdot a_2=48$

$\implies a_2 \cdot (1+r)=48$

$\boxed{\implies a_2=\frac{48}{1+r}}$

But ${\color{red}a_2 = r a_1}$.

Therefore ${\color{red}r a_1 = \frac{48}{1+r}}$

$\boxed{\color{red} \Rightarrow a_1 = \frac{48}{r(1+r)}}$ .... (2)

Substituting (2) into (1), we get:

$\boxed{{\color{red}\frac{48}{r(1+r)} = \frac{48}{1+r}} +32}$ .... (3)

Solve equation (3) for r. There are two solutions: r = 1/2 and r = -3. But only one of these will enable the infinite series to have a finite sum .....

The infinite sum is $S=a_1 \cdot \frac{1}{1-r}$ if $|r|<1$
I've done some editing (mostly in red) to Moo's fine reply. Since it's a diferent way (of getting the same thing) you might follow it a bit better ....

6. Thankssss a million!!!! I thought you might have been a little off, Moo, but thanks anyway you did help a lot. And thanks to Mr. Fantastic, now I understand completely. The only thing is, would the whole equation work if I substituted for example a(lower 1) with just a, and a(lower 2) with ar etc.???

Thanks again!

Steve

7. Originally Posted by Stevo_Evo_22
Thankssss a million!!!! I thought you might have been a little off, Moo, but thanks anyway you did help a lot. And thanks to Mr. Fantastic, now I understand completely. The only thing is, would the whole equation work if I substituted for example a(lower 1) with just a, and a(lower 2) with ar etc.???

Thanks again!

Steve
Yes. Personally I'd work with a1, a2 etc. But if you understand it better using a, ar, ar^2 etc. then you should do it that way.

8. Hi, sorry,

I'm stuck witht he same question. I end up with 2r^2 + 5r - 6.

I cannot factorise this for the life of me. The quadratic formula turns up weird answers, ditto for completing the square. Would I be able to get a bit of a hand with it?

Thanks

9. I'm sorry if my writing wasn't clear...

Originally Posted by Stevo_Evo_22
Hi, sorry,

I'm stuck witht he same question. I end up with 2r^2 + 5r - 6.

I cannot factorise this for the life of me. The quadratic formula turns up weird answers, ditto for completing the square. Would I be able to get a bit of a hand with it?

Thanks
Hm...There's a problem with your equation.

Let's do it :

$
\frac{48}{r(1+r)} = \frac{48}{1+r} +32
$

Multiplying both sides by r(1+r), we get :

$48=\frac{48 \cdot r \cdot (1+r)}{1+r}+32 \cdot r \cdot (1+r)$

$48=48r+32(r^2+r)$

$0=-48+48r+32r^2+32r$

$0=-48+80r+32r^2$

Dividing by 16 :

$2r^2+5r-{\color{red}3}=0$

Is it better ?

10. Originally Posted by Stevo_Evo_22
Hi, sorry,

I'm stuck witht he same question. I end up with 2r^2 + 5r - 6.

I cannot factorise this for the life of me. The quadratic formula turns up weird answers, ditto for completing the square. Would I be able to get a bit of a hand with it?

Thanks
$\frac{48}{r(1+r)} = \frac{48}{1+r} +32$

$\Rightarrow \frac{48}{r(1+r)} = \frac{48 r}{r(1+r)} + \frac{32r(1+r)}{r(1+r)}$

$\Rightarrow 48 = 48r + 32r(1+r)$

$\Rightarrow 32r^2 + 80r - 48 = 0$

$\Rightarrow 2r^2 + 5r - {\color{red}3} = 0$

$\Rightarrow (2r - 1)(r + 3) = 0$

etc.

11. AAAAHHHHHHHHH, thanks, I knew i did something wrong!!!!!

You've been a massive help to me, thanks a lot!

12. Hello, Stevo_Evo_22!

In a certain geometric sequence, the first term exceeds the second term by 32
and the sum of the second and third terms is 48.
Find the infinite sum of the sequence.
The first three terms are: . $a,\;ar,\;ar^2$

The first is 32 more than the second: . $a \:=\:ar-32\quad\Rightarrow\quad a \:=\:\frac{32}{1-r}\;\;{\color{blue}[1]}$

The sum of the 2nd and 3rd is 48: . $ar + ar^2 \:=\:48\quad\Rightarrow\quad ar(1+r) \:=\:48\;\;{\color{blue}[2]}$

Substitute [1] into [2]: . $\frac{32}{1-r}\cdot r(1+r) \:=\:48$

. . which simplifies to: . $2r^2 + 5r - 3 \:=\:0$

. . which factors: . $(2r-1)(r+3) \:=\:0$

. . and has roots: . $r \;=\;\frac{1}{2},\;-3$

Since a convergent series must have $|r| < 1$, the ratio is: . $r\:=\:\frac{1}{2}$

Substitute into [1]: . $a \:=\:\frac{32}{1-\frac{1}{2}} \quad\Rightarrow\quad a\:=\:64$

Therefore, the sum is: . $S \;=\;\frac{a}{1-r} \;=\;\frac{64}{1-\frac{1}{2}} \;=\;\boxed{128}$