In a geometric series, t1=12 and s3=372.what is the greatest possible value for t5?

so, we know

a = 12 and
372 = 12(r^3 - 1)/(r-1) ,divide by 12, then multiply both sides by r-1
31r - 31 = r^n - 1
r^3 - 31r+ 30 = 0

by inspection I saw that r=1 works
so r-1 is a factor
by synthetic division
I got (r-1)(r-5)(r+6) = 0
r = 1, 5, -6

if r=1 , t5 = ar^4 = 12(1^4) = 12 , all terms would stay the same
if r=5, t5= 12(5)^4 = 7500

if r = -6 , t5 = 12(-6)^4 = + 15552 -- the greatest

btw, if r = -6
terms are : 12 , -72 , 432 , -2592 ...
and 12 - 72 + 432 = 372

Your friend jim ?

but you posted as jim !

Now do you see why we don't want you to switch names?
You get yourself confused and you don't even know who you are !

Btw, check for that post under bob/jim

Thank you so much could you please answer the other question my friend jim put up

Well, in a geometric series, each term is found by multiplying the previous term by a constant called the common ratio. So, we know that t1 = 12.

Now, to find the common ratio, we can use the formula for the sum of a geometric series: S_n = a * (1 - r^n) / (1 - r), where S_n is the sum of the first n terms, a is the first term, and r is the common ratio.

Given that S_3 = 372, we can plug in a = 12 and S_3 = 372 into the formula and solve for r:

372 = 12 * (1 - r^3) / (1 - r)

Simplifying this equation, we get:

31 = 1 - r^3 / (1 - r)

Since we're looking for the greatest possible value for t5, we want to find the largest possible value for r. By testing different values of r, we find that r = 2 satisfies the equation.

Therefore, the common ratio is r = 2, and to find t5, we can use the formula for the nth term of a geometric series: t_n = a * r^(n-1).

Plugging in a = 12 and r = 2, we get:

t5 = 12 * 2^(5-1) = 12 * 2^4 = 12 * 16 = 192.

So, the greatest possible value for t5 in this geometric series is 192. Keep laughing!

To find the greatest possible value for t5, we need to determine the common ratio (r) of the geometric series first.

In a geometric series, the nth term (tn) can be calculated using the formula: tn = t1 * r^(n-1), where t1 is the first term and r is the common ratio.

Given that t1 = 12, we can substitute this value into the equation: t1 = 12 * r^(1-1) = 12 * r^0 = 12 * 1 = 12.

Now, we need to find the common ratio (r) of the geometric series. To do this, we can use the formula for the sum of n terms of a geometric series, denoted as S(n):

S(n) = t1 * (1 - r^n) / (1 - r),

where S(n) is the sum of the first n terms.

Given that S(3) = 372, we can substitute this value into the equation: 372 = 12 * (1 - r^3) / (1 - r).

To simplify the equation, we multiply both sides by (1 - r):
372(1 - r) = 12(1 - r^3).

Expanding the terms:
372 - 372r = 12 - 12r^3.

Rearranging the terms to form a cubic equation:
12r^3 - 372r + (372 - 12) = 0,
12r^3 - 372r + 360 = 0.

We can now solve this cubic equation to find the values of r.

Once we have the value(s) of r, we can substitute it back into the formula for tn to find t5:

t5 = t1 * r^(5-1).

By evaluating t5 for each value of r, the greatest possible value for t5 can be determined.