Arithmetic and geometric sequence

[paix1988]

*I am struggling with arithmetic and geometric sequences.

if the 4th term is m-8, 6th term 8m+3 and 8th term is 10m-5
Calculate the 1st and 5th term
Which term will have a value of -70The 4th term of geometric sequence is -16 and the 6th term is -64. Calculate the 3rd and 5th terms.

thank you for your assistance

 

[Greg]

Hello and welcome to MHB! paix1988 :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[MarkFL]

*I am struggling with arithmetic and geometric sequences.

if the 4th term is m-8, 6th term 8m+3 and 8th term is 10m-5
Calculate the 1st and 5th term
Which term will have a value of -70The 4th term of geometric sequence is -16 and the 6th term is -64. Calculate the 3rd and 5th terms.

thank you for your assistance

For the benefit of the community, I am going to post solutions.

1.) We are given the following information regarding an arithmetic progression:

\(\displaystyle a_4=m-8,\,a_6=8m+3,\,a_8=10m-5\)

If we denote the common difference as $d$, then from this we know:

\(\displaystyle 2d=(8m+3)-(m-8)=7m+11\)

\(\displaystyle 2d=(10m-5)-(8m+3)=2m-8\)

Equating the two results, we obtain:

\(\displaystyle 7m+11=2m-8\implies m=-\frac{19}{5}\implies d=-\frac{39}{5}\)

Now, in general, we have:

\(\displaystyle a_n=a_1+(n-1)d\)

And we know:

\(\displaystyle a_4=a_1+3d\)

\(\displaystyle m-8=a_1+3d\)

\(\displaystyle -\frac{59}{5}=a_1-\frac{117}{5}\)

\(\displaystyle a_1=\frac{58}{5}\)

Hence, the general term is:

\(\displaystyle a_n=\frac{58}{5}+(n-1)\left(-\frac{39}{5}\right)=\frac{58-39(n-1)}{5}\)

Thus, the 5th term is:

\(\displaystyle a_5=\frac{58-39(5-1)}{5}=-\frac{98}{5}\)

To find which term has a value of -70, we may write:

\(\displaystyle \frac{58-39(n-1)}{5}=-70\)

\(\displaystyle 58-39(n-1)=-350\)

\(\displaystyle -39(n-1)=-408\)

\(\displaystyle n-1=\frac{136}{13}\)

We will not get an integral value for $n$, thus none of the terms of this AP has a value of -70.

2.)The $n$th term of a GP is:

\(\displaystyle a_n=ar^n\)

We know that:

\(\displaystyle \frac{g_6}{g_4}=\frac{-64}{-16}=4=r^2\implies r=\pm2\implies a=-1\)

Thus, there are two possible GPs satisfying the given information:

a) \(\displaystyle a_n=-(-2)^n\implies a_3=8,\,a_5=32\)

b) \(\displaystyle a_n=-2^n\implies a_3=-8,\,a_5=-32\)