Sequence and Series (1 Viewer)

[boogerboy999]

Hey guys, i was wondering if you could help me out on this Q:

Given a geometric sequence whose sum of the first 10 terms is 4 and whose sum from the 11th to the 30th term is 48, find the sum from the 31st to the 60th term.

Im kinda stuck at the moment :S

 

[Timothy.Siu]

Hey guys, i was wondering if you could help me out on this Q:

Given a geometric sequence whose sum of the first 10 terms is 4 and whose sum from the 11th to the 30th term is 48, find the sum from the 31st to the 60th term.

Im kinda stuck at the moment :S

is it 1404?
i'm not sure if i did it correctly, it seemed pretty complicated...
i did the geometric sums for the first 10 terms and geometric sum of the first 30 terms in terms of A and R, then i managed to use them simulatenously to form a 3 degree polynomial and solve it...

 

[boogerboy999]

hmm
for the 1st ten numbers: S=a(r^10 -1)/(r-1)=4
for the 1st 30 numbers: S= a(r^30 -1)/(r-1)=48+4=52

so i divided them and got: (r^30-1)/(r^10-1)=13

now im stuck again xD

Edit: Ive got it solved:
Sum between 31-60= Sum60-Sum30
and work it out from there

The final answer i got was 52X1728=89856

Courtesy of JRAHS maths problems XD

 

[Trebla]

Hey guys, i was wondering if you could help me out on this Q:

Given a geometric sequence whose sum of the first 10 terms is 4 and whose sum from the 11th to the 30th term is 48, find the sum from the 31st to the 60th term.

Im kinda stuck at the moment :S

We want: S₆₀ - S₃₀

S₁₀ = 4
.: a(r¹⁰ - 1) / (r - 1) = 4

S₃₀ - S₁₀ = 48
=> S₃₀ = 52
.: a(r³⁰ - 1) / (r - 1) = 52

Dividing the two expressions:
(r³⁰ - 1) / (r¹⁰ - 1) = 52/4
(r¹⁰ - 1)(r²⁰ + r¹⁰ + 1) / (r¹⁰ - 1) = 13
r²⁰ + r¹⁰ + 1 = 13
r²⁰ + r¹⁰ - 12 = 0
(r¹⁰ + 4)(r¹⁰ - 3) = 0
r¹⁰ = 3 or r¹⁰ = - 4
Obviously r¹⁰ = 3 only because a real number raised to an even power is always positive

S₆₀ = a(r⁶⁰ - 1) / (r - 1)
= a(r³⁰ - 1)(r³⁰ + 1) / (r - 1)
= 52(r³⁰ + 1)
= 52((r¹⁰)³ + 1)
= 52((3)³ + 1)
= 1456

.: S₆₀ - S₃₀ = 1456 - 52 = 1404