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help with the hardest question in the universe (1 Viewer)
- Thread starter malcolm21
- Start date
Since you proved similar triangles earlier on
From (i)
q / 8 = 27 / p
(Corresponding sides of similar triangles are in proportion)
From (ii)
q / p = p / 8
(Corresponding sides of similar triangles are in proportion)
Then you show that there is a common ratio by using T2 / T1 = T3 / T2 or similar
From (i)
q / 8 = 27 / p
(Corresponding sides of similar triangles are in proportion)
From (ii)
q / p = p / 8
(Corresponding sides of similar triangles are in proportion)
Then you show that there is a common ratio by using T2 / T1 = T3 / T2 or similar
i dont get that last part?
T stands for term
By doing that you will find the common ratio (It's how to prove if something is a GP)
For an AP: T2 - T1 = T3 - T2
Bro. i know that. (obviously), and yes in a GP T1/T2 = Tn/T(n-1)
but when i got the similar triangle side ratios i didn't know what to do after that.
I solved them simultaenously to find p and q (yes i know it says hence but w/e, but what do i do to those ratios to prove its in a GP)
Haha. i didn't read the question properly...Show they are the FIRST FOUR.