help with the hardest question in the universe (1 Viewer)

[malcolm21]

literally combining my 2 worst topics

can anyone help me with part c & d?

 

[Crisium]

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

 

[Mr_Kap]

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

i dont get that last part?

 

[Crisium]

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

 

[Mr_Kap]

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)

 

[Mr_Kap]

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.