infinite geometric progressionns

[SabziiKumari]

hi i am a little stuck.

1) find the common ratio of geometric series.
2nd term = 6
sum to infinity = 24

******
for this is tried to create some equations
so we had ,,, ar = 6

then rearrange 6/1-r = 24 --> a = 24(1-r)
substitute that into ar=6

24r - 24r^2 = 6 ... neating it up you get .. 18=24r .. R = 18/24 = 3/4
the answere should b 1/2
where have i gone wrong.
thanks

2) a GP has a 2nd term of 6 and sum to infinity of 27. write down possible value of the first three terms. - i have no idea for this question.

 

[mmm4444bot]

ar = 6

then rearrange 6/(1 - r) = 24 We need to type parentheses around numerators and denominators in algebraic fractions!

The second equation above is not correct because the first term is not 6.

The first term is 6/r.

\(\displaystyle S \;=\; \frac{\frac{6}{r}}{1 - r}\)

?

2) a GP has a 2nd term of 6 and sum to infinity of 27. write down possible value of the first three terms. - i have no idea for this question.

I think we start this exercise the same way as the first one.

 

[Deleted member 4993]

hi i am a little stuck.

1) find the common ratio of geometric series.
2nd term = 6
sum to infinity = 24

******
for this is tried to create some equations
so we had ,,, ar = 6


The equation for sum of GP series is
\(\displaystyle S_n = \sum_{i=0}^n ar^i = \frac{a(1-r^{n+1})}{1-r}\)

so sum of infinite series when |r|<1
\(\displaystyle S_{\infty} = \frac{a}{1-r}\)

So we get

a = 24(1-r)

we know

ar = 6

then

24r(1-r) = 6

4r[sup:2ip3xtft]2[/sup:2ip3xtft] - 4r + 1 = 0 ? quadratic equation

(2r - 1)[sup:2ip3xtft]2[/sup:2ip3xtft] = 0 ? r = 1/2


then rearrange 6/(1-r) = 24 ?


--> a = 24(1-r)
substitute that into ar=6

24r - 24r^2 = 6 ... neating it up you get .. 18=24r .. R = 18/24 = 3/4
the answere should b 1/2
where have i gone wrong.
thanks

2) a GP has a 2nd term of 6 and sum to infinity of 27. write down possible value of the first three terms. - i have no idea for this question.

Follow the same route shown above ? find r, then find a and then find three terms.

 

[soroban]

Hello, SabziiKumari!

1) Find the common ratio of this geometric series:
. . 2nd term = 6, .sum to infinity = 24

\(\displaystyle \text{The first term is: }a.\)
\(\displaystyle \text{The second term is: }ar\!:\;\;ar \:=\:6 \quad\Rightarrow\quad a \:=\:\frac{6}{r}\;\;[1]\)
\(\displaystyle \text{The sum to infinity is 24: }\;\frac{a}{1-r} \:=\:24 \quad\Rightarrow\quad a \:=\:24 - 24a \;\;[2]\)


\(\displaystyle \text{Equate [1] and [2]: }\;\frac{6}{r} \:=\:24 - 24r \quad\Rightarrow\quad 6 \:=\:24r - 24r^2 \quad\Rightarrow\quad 4r^2 - 4r + 1 \:=\:0\)

\(\displaystyle \text{Therefore: }\2r-1)^2 \:=\:0 \quad\Rightarrow\quad r \:=\:\frac{1}{2}\)

2) A GP has a 2nd term of 6 and sum to infinity of 27.
Find the possible values of the first three terms.

We use the same approach.

\(\displaystyle \text{The second term is }ar\!:\;\;ar \:=\:6 \quad\Rightarrow\quad a \:=\:\frac{6}{r}\;\;[1]\)

\(\displaystyle \text{The sum to infinity is 27: }\:\frac{a}{1-r} \:=\:27 \quad\Rightarrow\quad a \:=\:27 - 27r\;\;[2]\)

\(\displaystyle \text{Equate [1] and [2]: }\;\;\frac{6}{r} \:=\:27-27r \quad\Rightarrow\quad 6 \:=\:27r - 27r^2 \quad\Rightarrow\quad 9r^2 - 9r + 2 \:=\:0\)

\(\displaystyle \text{Hence: }\3r-1)(3r-2) \:=\:0 \quad\Rightarrow\quad r \:=\:\frac{1}{3},\:\frac{2}{3}\)


\(\displaystyle \text{If }r=\frac{1}{3}\text{, then }a = 18\)

. . \(\displaystyle \text{The first three terms are: }\:18,\:6,\:2\)


\(\displaystyle \text{If }r = \frac{2}{3}\text{, then }a = 9\)

. . \(\displaystyle \text{The first three terms are: }\:9,\:6,\:4\)