geometric progression question please

[val1]

I really need help with this one:

The second and fifth terms in a geometric progression are 405 and -120 respectively. Find the seventh term and the sum of the first seven terms.


The formulas for finding the sum of the terms and for finding the nth term both require knowing the first term, a, and the common ratio, r, neither of which are known.

How do I solve this question?

Thanks

 

[galactus]

You are given \(\displaystyle a_{2}=405, a_{5}=-120\)

Use the formula \(\displaystyle a_{n}=a_{1}r^{n-1}\)

We can build a system of equations and solve:

\(\displaystyle a_{2}=a_{1}r^{1}\)

\(\displaystyle a_{5}=a_{1}r^{4}\)


\(\displaystyle 405=a_{1}r^{1}\)

\(\displaystyle (-120)=a_{1}r^{4}\)


Now solve for \(\displaystyle a_{1}\) in the first equation and sub into the second, solve for r.

Then, to find the sum, you can use the formula:

\(\displaystyle a_{1}(\frac{1-r^{n}}{1-r})\)

 

[Guest]

GP's

form is
the n th a term = a (r ^(n-1))

second term = 405

405 = a ( r ^(2-1))

405 = a . r

and the second equation is...

-120 = a( r ^ ( 5-1))

-120 = a ( r ^4) )

now solve these 2 equations.