find the sum of the 15 terms in the geometric sequence 4,12,36,108

Question

raden · Answer

use formula :
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sn = a1(r^n-1)/(r-1)
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with : a1=1st term = 4
r=ratio=a2/a1=12/4=3
and just put n=15

anonymous · Answer

Geometric series have a common multiplier between each sequential number. notice how 12/4=36/12=108/36=3
Find the ratio and the summation is just your first term times your ratio raised the number of terms you have.
$$S=a*r^n$$ where a is your first term, r is your ratio and n is the number of terms you have.

anonymous · Answer

i was wrong it should be $$S=a*r^{n-1}$$

raden · Answer

dont forget, divided by (r-1)

anonymous · Answer

if you divide by r-1, you get a wrong answer. try the first term with your equation. 4*(4^(1-1))/(3-1)=4*1/2=2 The first term is clearly 4 as the sequence shows but your /(r-1) reduces that by half. That half carries through the sequence making all summations reduced by 1/2

raden · Answer

re-check that question...
we find the sum of the 15 terms, not find the 15th term

anonymous · Answer

ah you're correct, my bad. i was mistaken.

anonymous · Answer

thank you so much

raden · Answer

welcome