Geometric Progression

[rubbishatmaths121]

Hi All,

I am stuck on a question on my paper. Can anyone help? and show me your workings so I can follow and learn?


For the following geometric progression 1, 1/2, 1/4 ........ determine

· The 20th term of the progression

· The value of the sum when the number of terms in the sequence tends to infinity and explain why the sequence tends to this value Sn = ∑n=0 n→∞ arn

 

[MathLover1]

For the following geometric progression 1, 1/2, 1/4 ........ determine

nth term formula of geometric sequence is:

a[n]=a[1]*r^(n-1) where a[1] is first term and r is ratio

given:
a[1]=1
a[2]=1/2

ratio is (1/2)/1=1/2

then nth term formula is:

a[n]=1*(1/2)^(n-1)


The 20th term of the progression will be:

20th term=> n=20

a[20]=1*(1/2)^(20-1)
a[20]=1*(1/2)^19
a[20]=1/524288

·The value of the sum when the number of terms in the sequence tends to infinity and explain why the sequence tends to this value Sn = ∑n=0 n→∞ arn

As you can see, 20th term has large denominator. As you go with more terms denominator will be increasingly larger each time until reaches infinity

and

so, your sequence

this geometric series converge (have a limit) :

which explains why the sequence tends to this value Sn = ∑n=0 n→∞

 

[nycmathguy]

For the following geometric progression 1, 1/2, 1/4 ........ determine

nth term formula of geometric sequence is:

a[n]=a[1]*r^(n-1) where a[1] is first term and r is ratio

given:
a[1]=1
a[2]=1/2

ratio is (1/2)/1=1/2

then nth term formula is:

a[n]=1*(1/2)^(n-1)


The 20th term of the progression will be:

20th term=> n=20

a[20]=1*(1/2)^(20-1)
a[20]=1*(1/2)^19
a[20]=1/524288

·The value of the sum when the number of terms in the sequence tends to infinity and explain why the sequence tends to this value Sn = ∑n=0 n→∞ arn

As you can see, 20th term has large denominator. As you go with more terms denominator will be increasingly larger each time until reaches infinity

and

so, your sequence

this geometric series converge (have a limit) :

which explains why the sequence tends to this value Sn = ∑n=0 n→∞

Very impressive work. Is this a precalculus topic?

 

[MathLover1]

yes

 

[nycmathguy]

yes

Ok. I can't wait to get there.