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

Question

anonymous · Answer

is it the a+ar+ar^2...  formula?

anonymous · Answer

a+ar+ar^2  is short form for 
ar^0+ar^1+ar^2 ....   ar^7
 1        2        3           8th will be

anonymous · Answer

a is just 8 since it is first term
a=8 
r is    
ar^7=781250   ->    8*r^7=781250  solve for r

anonymous · Answer

r=5.162

anonymous · Answer

knowing that you then use this formula http://www.regentsprep.org/Regents/math/algtrig/ATP2/GeoSeq.htm to find the sum :D

anonymous · Answer

(8*(1-5.162^8)/(1-5.162)

anonymous · Answer

roughly 
969016.23049467104062767206

campbell_st · Answer

just a slight correction to the above work 

$$a_{n} = a_{1} \times r^{n -1}$$

you know a1 = 10 and an = 781250 and n = 8

so you have 

$$781250 = 10 \times r^{8 -1}$$

divide both sides of the equation by 10

$$78125 = r^7$$

to find the common ratio r 

$$\sqrt[7]{78125} = r$$

so in actual fact r = 5

for the sum of a geometric series

$$S_{8} = \frac{10(1 - 5^8)}{1 - 5}$$

evaluate for the answer. 

Hope this helps.