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If-a-b-c-are-in-GP-and-log-c-a-log-b-c-log-a-b-are-in-AP-then-the-common-difference-of-the-AP-is-




Question Number 43602 by peter frank last updated on 12/Sep/18
If  a, b, c  are in GP and log_c a, log_b c, log_a b  are in AP, then the common difference  of the AP is
Ifa,b,careinGPandlogca,logbc,logabareinAP,thenthecommondifferenceoftheAPis
Answered by $@ty@m last updated on 12/Sep/18
Let log _c a=x−y ⇒a=c^(x−y)  ...(1)  log _b c=x  ⇒b^x =c ...(2)  log _a b=x+y ⇒a^(x+y) =b ...(3)  From (1) &(2),  a=b^(x(x−y))   ...(4)  From (3) &(4),  a=a^((x+y)x(x−y))   ⇒(x+y)x(x−y)=1 ...(5)  ∵ a, b & c are in G.P.  ∴ b^2 =ac  ⇒a^(2(x+y)) =a.b^x from (2)&(3),  ⇒a^(2(x+y)) =a.a^(x+y)   ⇒a^(x+y) =a  ⇒x+y=1  ⇒x=1−y  Substituting this in (5)  x(x−y)=1  ⇒(1−y)(1−2y)=1  ⇒1−3y+2y^2 =1  ⇒y(2y−3)=0  ⇒y=0 (inadmissible)  or y=(3/2) Ans.
Letlogca=xya=cxy(1)logbc=xbx=c(2)logab=x+yax+y=b(3)From(1)&(2),a=bx(xy)(4)From(3)&(4),a=a(x+y)x(xy)(x+y)x(xy)=1(5)a,b&careinG.P.b2=aca2(x+y)=a.bxfrom(2)&(3),a2(x+y)=a.ax+yax+y=ax+y=1x=1ySubstitutingthisin(5)x(xy)=1(1y)(12y)=113y+2y2=1y(2y3)=0y=0(inadmissible)ory=32Ans.

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