Question Number 10318 by Tawakalitu ayo mi last updated on 03/Feb/17

If 12, x, y and 4 provides a sequence such that  the first 3 of the numbers are in arithmetic  progression. Calculate the   (a) Possible values of x and y  (b) The sum of the A.P  (c) The product of the last 3 numbers of  the G.P

Answered by mrW1 last updated on 05/Feb/17

(a)  the first 3 numbers are in A.P.  ⇒ y−x=x−12     (i)  the last 3 numbers are in G.P.:  ⇒(4/y)=(y/x)     (ii)    from (i) we get:  y=2x−12=2(x−6)    from (ii) we get:  y^2 =4x  4(x^2 −12x+36)=4x  x^2 −13x+36=0  x=((13±(√(13^2 −4×36)))/2)=((13±5)/2)= { (9),(4) :}  y=2(x−6)= { (6),((−4)) :}    the sequence is  12, 9, 6, 4 or  12, 4, −4, 4    (b)  12+9+6=27 or  12+4−4=12    (c)  9×6×4=216 or  4×(−4)×4=−64

Commented byTawakalitu ayo mi last updated on 04/Feb/17

I really appreciate your effort sir. God bless  you sir.