At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

i dont think it uploaded properly try again and ill try and help

i'll draw it

ok :)

sorry i didnt read

i think its a determinant

what does GP stand for?

geometric series

ah nice. ok give me a minute

i assume that they are consecutive terms - can we do that?

unfortunately not, but we do know their positions, p,q,r

the nth term of a GP is A(r^(n-1)) where A is the first term and r is the common ratio

to avoid confusion lets use d instead of r

so lets call the first term of the progression "f" and the common ratio "d"

can you see this:
\[a= fd^{p-1}\]

yes

cool have a go playing with that, and see how far you can get it

b = fd^(q-1)

yes

hows it going?

i'm struggling - whats confusing me is that it says do not expand

ah ok well ill put up my solution and you can read through as far as you like

ok

\[\left|\begin{matrix}log(a) & p & 1\\ log(b) & q & 1\\ log(c) & r & 1\end{matrix}\right| \]

now just to check, are you familiar with how you can manipulate determinants, eg row operations?

yes - i got that far! - no more!

only vaguely

thats why I'm stuck

ok

yea i follow that

oh yes i remember doing something like this - to find inverse of matrix i think

i think i know whats coming - when you calculate the determinant the zeros will be significant

to be honest no

ok no worries, ill explain :)

yes i think i see now - those elements in row 3 are multiplied by 0

there are two ways to think about it, by expanding or by operations, whichever you prefer :)

does this help or is it still confusing? i hope i haven't made it worse lol

here is a good site http://www.sosmath.com/matrix/determ1/determ1.html

thanks a lot

no problem :) thanks for the juicy problem