Question Number 103780 by bobhans last updated on 17/Jul/20

the third, sixth and seventh terms of a  geometric progression (whose common  ratio is neither 0 nor 1 ) are in  arithmetic progression . prove that the  sum of the first three terms is equal to  fourth .

Answered by bemath last updated on 17/Jul/20

AP : ar^2  , ar^5 , ar^6  ⇒2ar^5  = ar^2 +ar^6   2r^5  = r^2 +r^6  ⇒ 2r^3  = 1+r^4   r^4 −2r^3 +1 = 0 ; (r−1)(r^3 −r^2 −r−1)=0  r^3 = r^2 +r + 1  now we have T_4 =ar^3   sum of the first three terms is   a+ar+ar^2  = a(1+r+r^2 )   = a(r^3 ) = T_4  . proved ■