Geometric progression

**loui1410** · January 8th 2008, 12:00 PM

Hello guys, I need help solving the following question:

The sum of the first three terms in a geometric progression is 195. The difference (?) between the first and second term is larger in 75 than the third term. Find the first term and the common ratio of the given progression.

That is:
I a1 + a2 + a3 = 195
II a2 - a1 = a3 + 75
a1=? q=?

Could I please have the full solution? (because I reached a stage where I couldn't continue and didn't know what to do next)

Thanks in advance.

**topsquark** · January 8th 2008, 12:12 PM

Originally Posted by loui1410

Hello guys, I need help solving the following question:

The sum of the first three terms in a geometric progression is 195. The difference (?) between the first and second term is larger in 75 than the third term. Find the first term and the common ratio of the given progression.

That is:
I a1 + a2 + a3 = 195
II a2 - a1 = a3 + 75
a1=? q=?

Could I please have the full solution? (because I reached a stage where I couldn't continue and didn't know what to do next)

Thanks in advance.

Define
$a_n = a_1r^{n - 1}$
where r is the geometric ratio.

So
$a_2 = a_1r$
and
$a_3 = a_1r^2$

So your conditions are
$a_1 + a_1 r + a_1 r^2 = 195$
and
$a_1 r - a_1 = a_1 r^2 + 75$

Thus we have two equations in two unknowns.

The first condition reads:
$a_1(r^2 + r + 1) = 195$
and the second reads
$a_1(-r^2 + r - 1) = 75$

There are a variety of ways to attack this. This is one:
Solve the top equation for $a_1$ :
$a_1 = \frac{195}{r^2 + r + 1}$

and insert it into the bottom equation:
$\left ( \frac{195}{r^2 + r + 1} \right )(-r^2 + r - 1) = 75$

And finally, multiply both sides by $r^2 + r + 1$ :
$195(-r^2 + r - 1) = 75(r^2 + r + 1)$

This is a quadratic in r, which I am sure you can solve.

Note: For good or for ill I get that r is a complex number. I see no reason why this shouldn't be so, but I doubt you expected it. As I am not familiar with seeing problems like this turn out to be complex, I am wondering if there isn't some kind of typo?

-Dan

**loui1410** · January 8th 2008, 12:42 PM

Thanks a lot for your help. No there's no typo, the r is supposed to have two solutions: 1/3 and -4/3. But the problem is that the discriminant of the quadratic equation is negative!

$-195r^2+195r-195=75r^2+75r+75$
$270r^2-120r+270=0$
Δ $=144-4*27*27=-2772$

Where's my mistake? I can't find it!

**mr fantastic** · January 8th 2008, 02:01 PM

Originally Posted by loui1410

Thanks a lot for your help. No there's no typo, the r is supposed to have two solutions: 1/3 and -4/3. But the problem is that the discriminant of the quadratic equation is negative!

$-195r^2+195r-195=75r^2+75r+75$
$270r^2-120r+270=0$
Δ $=144-4*27*27=-2772$

Where's my mistake? I can't find it!

There's no mistake. Topsquark has already implied this technical infelicity:

Originally Posted by topsquark

[snip]
Note: For good or for ill I get that r is a complex number. I see no reason why this shouldn't be so, but I doubt you expected it. As I am not familiar with seeing problems like this turn out to be complex, I am wondering if there isn't some kind of typo?

-Dan

which is why he asked if there's a typo.

If there's no typo in what you posted, then the typo must be in the source of the question (textbook? problem sheet from class? Such things aren't infallible ......)

Alternatively, Topsquark's (quite reasonable) mathematical interpretation of

"The difference (?) between the first and second term is larger in 75 than the third term."

could be different to what the source of the question intends .....

**loui1410** · January 8th 2008, 02:27 PM

It is supposed to be $a_1-a_2=a_3+75$ , not $a_2-a_1=a_3+75$ as I thought. The language which was the question originally written in is not my mother-tongue, and English isn't my mother-tongue neither. So sorry guys, it was only a result of mistranslation

Anyway, I reached the correct solution now.

Thanks again.

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