What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is
781,250?

Question

What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 
781,250?

anonymous · Answer

how am i supposed to answer this if i dont know the common ratio ?

whpalmer4 · Answer

If you have two terms in the sequence, $n_1$ and $n_2$, you know that the common ratio is $a = \frac{n_2}{n_1}$ right?

anonymous · Answer

noo, so its 8/10 ?

whpalmer4 · Answer

What if you had two terms, but they were $n_1$ and $n_3$?  Then you know that $n_2=n_1*a$ and $n_3= n_2*a$ so $n_3=(n_1*a)*a = n_1*a^2$

whpalmer4 · Answer

If you knew the 1st and 4th terms, then $n_4 = n_1*a*a*a = n_1*a^3$ can you see the pattern developing here?

anonymous · Answer

this is confusing :(. is that how to find the common ratio ?

whpalmer4 · Answer

Yes.  If you know terms 1 and n of a geometric sequence, then term n = term 1 * common ratio to the (n-1) power.  Let's try an example.  Say the common ratio is 2.  We know that the first term is 3, and we want to find the 3rd term.  Well, that's going to be 3*2*2 = 12, right?  Suppose we instead knew that we had a 3 term sequence, the first term was 3, and the 3rd term was 12.  We could find the common ratio by dividing 12/3 = 4.  4 represents the product of all of the common factor multiplications to go from the 1st term to the 3rd term.  That's 2 terms, so 4 must be the common ratio squared, right?  and indeed, 4 = 2*2

whpalmer4 · Answer

For your problem, we have 8 terms, the first is 10, and the 8th is 781250.  If we divide the 8th by the 1st, we get 78125.  Our common ratio to the 7th power (8-1) = 78125.  We could just try some numbers: 2^7 = 128, not big enough.  10^7 = 10000000, too big.  How about 5^7?  What do you know, 5^7 = 78125!  So our common ratio is 5.

whpalmer4 · Answer

Given that the ratio ended in 5, 5 was actually a pretty obvious candidate to try first, as all numbers ending in 5 or 0 are divisible by 5...

anonymous · Answer

This is soooo hard :/

whpalmer4 · Answer

It's a bit tricky at first, I agree.

anonymous · Answer

for geometric sequence a term is given by
T=a(r^n-1)
where n is the number of term to be found
in this case for last term T=781250 n=8 a=10 plug in values and solve for r

anonymous · Answer

and its T=a(r^(n-1))

whpalmer4 · Answer

@lesha here's what you see for the first 8 terms for common ratios from 2 through 7:
2: {10, 20, 40, 80, 160, 320, 640, 1280}
3: {10, 30, 90, 270, 810, 2430, 7290, 21870}
4: {10, 40, 160, 640, 2560, 10240, 40960, 163840}
5: {10, 50, 250, 1250, 6250, 31250, 156250, 781250}
6: {10, 60, 360, 2160, 12960, 77760, 466560, 2799360}
7: {10, 70, 490, 3430, 24010, 168070, 1176490, 8235430}

whpalmer4 · Answer

You can see that a common ratio of 5 is the one that gets you from 10 to 781250 in 8 terms.