SuicideFuel Problem

[trying to ascend]

[UWSL]Between the numbers 3 and 192 are inserted equal numbers of terms of an arithmetic progression and of a geometric progression with ratio r and q, respectively, where r and q are integers. Number 3 and number 192 participate in these two progressions. It is known that the third term of ,

(1 + 1/q) to the power of 8, in crescent powers of 1/q, equals r/9q. Therefore the second term of the AP is?[/UWSL]

 

[TheProphetMuscle]

Yes

 

[idk125]

very hard question ngl

 

[Caesercel]

66 ?

 

[trying to ascend]

66 ?

Correct!

What was your solution method?

 

[idk125]

66 ?

wow very smart mod

 

[Caesercel]

Correct!

What was your solution method?

Its a weirdly worded question. I had to throw my head at it a bit.

Third term of (1+1/q) ^8 = 28/q2. From that we get r*q = 28*9

Now here's the crux. There is some value of q that could be multiplied to 3 multiple times to get 192. I factorized 192 and it became obvious that q is either 2,4 or 8. So I tried all of them. Cannot be 8 otherwise r won't be integer. Cannot be 2 because the r we get does not lead from 3 to 192 in AP. So it has to be 4 . The r we get is 63 . Adding to 3,the second term in AP becomes 66.

You can check that 4 is multiplied 3 times to 3 to get 192 and 63 is added 3 times to 3 to get 192. So the answer checks out