Series and Sequences question HELP! (1 Viewer)

[star*eyed]

FInd the least number of terms for which the series 20 + 4 + 4/5... is greater than 24.99.

Can anyone answer that??

 

[star*eyed]

ANother question: The sum of the first 5 terms of a geometric series is 77 and the sim of the next 5 is -2464. Find the 4th term.

Geez i suck at this stuff

 

[Loz_metalhead]

FInd the least number of terms for which the series 20 + 4 + 4/5... is greater than 24.99.

Can anyone answer that??

Is the answer 2? If it is I will tell you how I worked it out

 

[SoulSearcher]

FInd the least number of terms for which the series 20 + 4 + 4/5... is greater than 24.99.

Can anyone answer that??

You have a = 20, r = 1/5 = 0.2

Therefore 24.99 < 20(1-0.2ⁿ)/(1-0.2)
99.96 < 100(1-0.2ⁿ)
0.9996 < 1-0.2ⁿ
0.2ⁿ < 0.0004
ln (0.2ⁿ) > ln 0.0004
n ln 0.2 > ln 0.0004
n > ln 0.0004 / ln 0.2
n > 4.86
Therefore n = 5 for the the sum to first go above 24.99

 

[pLuvia]

First one
Use the sum of a GP, S_n=a(1-rⁿ)/(1-r)
where
a=20
r=1/5
And let S_n>24.99

Second question
Similar to question one use the sum of a GP formula, find a and r, by using the two equations you get from the given information that is:
S₅=a(1-r⁵)/(1-r)
S₁₀=a(1-r¹⁰)/(1-r)
Solve simulatenously, and get your a and r values. Then use t_n=ar^n-1 and sub n=4 and you should get your answer

SS solved it for you

 

[star*eyed]

Is the answer 2? If it is I will tell you how I worked it out

Um i dunno wat the answer is....but cant it not be 2 caus isnt the answer to that 4??? i dunno though....

 

[Loz_metalhead]

Yeh. Oops. I'm not good at this either.

 

[SoulSearcher]

ANother question: The sum of the first 5 terms of a geometric series is 77 and the sim of the next 5 is -2464. Find the 4th term.

Geez i suck at this stuff

S₅ = a(r⁵-1)/(r-1) = 77 ... (1)
S₁₀ - S₅ = a(r¹⁰-1)/(r-1) - a(r⁵-1)/(r-1) = -2464 ... (2)
a(r¹⁰-1)/(r-1) - 77 = -2464
a(r¹⁰-1)/(r-1) = -2387 ... (3)
(3) / (1)
(r¹⁰-1)/(r⁵-1) = -2387/77
(r⁵-1)(r⁵+1)/(r⁵-1) = -31
r⁵+1 = -31
r⁵ = -32
Therefore r = -2
Therefore, substituting r = -2 into (1),
a((-2)⁵-1)/(-2-1) = 77
-33a = -231
a = 7
Therefore, fourth term is ar³
= 7 * (-2)³
= 7 * -8
= -56