I need help with this one too
Find the Sum the 4 terms
10, 30, 90, 270, ...
The second term of a geometric sequence is 9 and the fourth term is 81. Find the sum of the first 10 terms of this sequence.
How can I find the answer?
5 answers
r=3
nth term= 3^n
sn= sum from n=1 to 10
Sn= 3+3^2 + 3^3 + ...+3^10
http://www.dummies.com/education/math/calculus/how-to-find-the-partial-sum-of-a-geometric-sequence/
On the second, you have the four terms, add them.
nth term= 3^n
sn= sum from n=1 to 10
Sn= 3+3^2 + 3^3 + ...+3^10
http://www.dummies.com/education/math/calculus/how-to-find-the-partial-sum-of-a-geometric-sequence/
On the second, you have the four terms, add them.
I dont understand
You must know the basic formulas for a geometric sequence.
term(n) = a r^(n-1), where a is the first term, r is the common ratio and n is the term number
for yours:
ar = 9 and ar^3 = 81
divide one equation by the other:
ar^3/(ar) = 81/9
r^2 = 9
r = ± 3
then ar = 9
So if r = 3, a = 3, if r = -3, a=-3
sum(n) = a(r^n - 1)/(r-1)
case1:
Sum(10) = 3(3^10 - 1)/(3-1)
= 3( 59048)/2 = 88572
case2:
sum(10) = -3( (-3)^10 - 1)/(-3-1) = 44286
checking the 2nd:
series = -3 + 9 - 27 + 81 - 243 + 729 - 2187 + 6561 - 19683 + 59049
= 44286
For the second question, why not just add them up?
You only have 4 terms!
otherwise: a = 10, r = 3
sum(4) = 10(3^4 - 1)/(3-1) = 400
term(n) = a r^(n-1), where a is the first term, r is the common ratio and n is the term number
for yours:
ar = 9 and ar^3 = 81
divide one equation by the other:
ar^3/(ar) = 81/9
r^2 = 9
r = ± 3
then ar = 9
So if r = 3, a = 3, if r = -3, a=-3
sum(n) = a(r^n - 1)/(r-1)
case1:
Sum(10) = 3(3^10 - 1)/(3-1)
= 3( 59048)/2 = 88572
case2:
sum(10) = -3( (-3)^10 - 1)/(-3-1) = 44286
checking the 2nd:
series = -3 + 9 - 27 + 81 - 243 + 729 - 2187 + 6561 - 19683 + 59049
= 44286
For the second question, why not just add them up?
You only have 4 terms!
otherwise: a = 10, r = 3
sum(4) = 10(3^4 - 1)/(3-1) = 400
Got it thanks so much :,)
2 answers