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geometric progession help

first two terms of geometric progression are 10 and 4
i. find the sum of the first 15 terms
ii. find the sum to infinity in exact form

so sn=a(r^n-1)/r-1 a=10 n=15 and r=0.4

s15=10(0.4^14)/0.4-1= -4.473924267*10^-2 is this right?

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Reply 1
No it's not.

You seem to be missing a -1 in the numerator and it's to the power 15.
Reply 2
it is rn1r^n-1
Reply 3
Original post by dongonaeatu
first two terms of geometric progression are 10 and 4
i. find the sum of the first 15 terms
ii. find the sum to infinity in exact form

so sn=a(r^n-1)/r-1 a=10 n=15 and r=0.4

s15=10(0.4^14)/0.4-1= -4.473924267*10^-2 is this right?


No.

sn=arn1r1s15=100.41510.41\displaystyle s_n=a\frac{r^n-1}{r-1} \rightarrow s_{15}=10\cdot \frac{0.4^{15}-1}{0.4-1}
and the sum of infinity for |r|<1

s=a11r\displaystyle s=a\frac{1}{1-r}
(edited 11 years ago)
Reply 4
Original post by dongonaeatu
first two terms of geometric progression are 10 and 4
i. find the sum of the first 15 terms
ii. find the sum to infinity in exact form

so sn=a(r^n-1)/r-1 a=10 n=15 and r=0.4

s15=10(0.4^14)/0.4-1= -4.473924267*10^-2 is this right?


The formula for sum is, Sn=a(rn1)r1 \displaystyle S_n =\frac{a(r^{n}-1)}{r-1}

So inputting the values gives,
S15=10(0.4151)0.41 \displaystyle S_{15}=\frac{10(0.4^{15}-1)}{0.4-1}
Reply 5
Original post by ztibor
No.

sn=arn1r1\displaystyle s_n=a\frac{r^n-1}{r-1}
and the sum of infinity for |r|<1

s=a11r\displaystyle s=a\frac{1}{1-r}


i got 1.666638771 but i guess thats wrong too as its the sum of the first 15 terms and the first term is 10
Reply 6
Original post by raheem94
The formula for sum is, Sn=a(rn1)r1 \displaystyle S_n =\frac{a(r^{n}-1)}{r-1}

So inputting the values gives,
S15=10(0.4151)0.41 \displaystyle S_{15}=\frac{10(0.4^{15}-1)}{0.4-1}


raheem you tank

is the answer 16.66664877
Reply 7
Original post by ztibor
No.

sn=arn1r1s15=100.41510.41\displaystyle s_n=a\frac{r^n-1}{r-1} \rightarrow s_{15}=10\cdot \frac{0.4^{15}-1}{0.4-1}
and the sum of infinity for |r|<1

s=a11r\displaystyle s=a\frac{1}{1-r}


thats wrong the a should be above the division line
Reply 8
Original post by dongonaeatu
raheem you tank

is the answer 16.66664877


Yes
Reply 9
Original post by dongonaeatu
thats wrong the a should be above the division line


It doesn't matters, ztibor is correct.
Reply 10
Original post by raheem94
It doesn't matters, ztibor is correct.


ah okay. What is the formula for the sum to infinity
Original post by dongonaeatu
ah okay. What is the formula for the sum to infinity


ztibor has given you that
Reply 12
Original post by ztibor
No.

sn=arn1r1s15=100.41510.41\displaystyle s_n=a\frac{r^n-1}{r-1} \rightarrow s_{15}=10\cdot \frac{0.4^{15}-1}{0.4-1}
and the sum of infinity for |r|<1

s=a11r\displaystyle s=a\frac{1}{1-r}


it says find the sum to infinity in exact form so.

sum to infinity= 10/1-0.4= 16.6?
Reply 13
Original post by TenOfThem
ztibor has given you that


is the sum to infinity 16.6
10/0.6 = 100/6 = 50/3

you seem to have rounded so that would not be exact form
Reply 15
Original post by raheem94
It doesn't matters, ztibor is correct.


is the sum to infinity 16.6
Reply 16
Original post by dongonaeatu
is the sum to infinity 16.6


Yes, it is 16.7(to 3s.f.).
Reply 17
Original post by TenOfThem
10/0.6 = 100/6 = 50/3

you seem to have rounded so that would not be exact form


yeah because it had lots of 6's in it lol. can i just leave it as 50/3 then?
Reply 18
Original post by dongonaeatu
yeah because it had lots of 6's in it lol. can i just leave it as 50/3 then?


The question asks for exact answer hence you MUST write it as 503 or 1623 \displaystyle \frac{50}3 \ or \ 16\frac23
Reply 19
Original post by raheem94
The question asks for exact answer hence you MUST write it as 503 or 1623 \displaystyle \frac{50}3 \ or \ 16\frac23


hey man, solve 3*5^x=150 giving it to 4dp

is it; log3*5^x=log150

5^x=log150/log3

5^x=4.560876795 then i dont know how to get the x on its own

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