Question

Hi! Is anyone able to explain to me how to get the Number of Terms in a Geometric Progression when I only have the last value:

For example: GP: 2, 2*Sqrt(3), 6, '.......', 486

I know the formulas but i can't seem to do this problem involving unknown N. Please help

Answer 1

formula for General term for GP:

Answer 2

and the sum:

Answer 3

so I need to get the sum of the progression, but I don't know what term is 486 or how to get it :(

Answer 4

\[Find the \sum of GP: 2, 2\sqrt{3}, 6, ".....", 486\]

Answer 5

Can you use logarithms (in your class)?

Answer 6

hmm Logarithm is our next topic so I'm guessing I need that to solve this problem?

Answer 7

hint : 3^5 = 243

Answer 8

I think the way of @ganeshie8 is better.

Answer 9

so from 486 = 2 * 1.73^n-2

it becomes: 243 = 1.73^n-1

Answer 10

sorry n-2 is typo

Answer 11

Exact.

Answer 12

then (without knowing Logs) i would pretty much have to guess what 1.73 exponent is to amount to 243 ?

Answer 13

No needed. You can try this,
\[\sqrt{3}=3^{1/2}\]

Answer 14

Your equation should look like this,
\[3^5=3^{\frac{n-1}{2}}\]

Answer 15

sorry I'm a bit slow where did the 3s come from :(

Answer 16

You can do the decomposition in prime factos,
\[243=3^5\]

Answer 17

The 3 under the square root is the one you give in your original problem, I mean.

Answer 18

Do you understand it?

Answer 19

i understand the 243=3^5 bit

Answer 20

Now, in the original problem it was,
\[486=2(\sqrt{3})^{n-1}\]
You said correctly,
\[243=(\sqrt{3})^{n-1}\]And using decomposition,
\[3^5=(\sqrt{3})^{n-1}\]And now, we will use,
\[\sqrt{3}=3^{1/2}\]so,
\[3^5=(3^{1/2})^{n-1}\Rightarrow 3^5=3^{(n-1)/2}\]
Do you understand all steps and know how to solve?

Answer 21

ahhh ok i see it now thank you!!! :)))

Answer 22

You're welcome.