Given `a ,b ,c` are in A.P. `b ,c , d` are in G.P. and `c ,d ,e` are in H.P. If `a=2a d ne=18 ,` then the sum of all possible value of `c` is ________.

Explanation

Given,`a,b,c` are in AP.
`therefore " " b=(a+c)/(2) " " "…….(i)"`
`b,c,d` are in GP.
`therefore c^(2)=bd " " "…….(ii)"`
and `c,,e` are in HP.
`therefore d=(2ce)/(c+e) " " "…..(iii)"`
Now, substituting the values of b and d in Eq. (ii), than
` c^(2)= ((a+c)/(2))((2ce)/(c+e))`
`implies c(c+e)=e(a+c)`
`implies c^(2)=ae " " ".......(iv)"`
Given, `a=2,e=18`
From Eq. (iv), `c^(2)=(2)(18)=36`
` therefore " " c= pm 6 `
From Eq. (i), `b=(2 pm 6)/(2)=4,-2`
and from Eq. (ii), `d=(c^(2))/(b)=(36)/(b)=(36)/(4) " or " (36)/(-2)`
`therefore " " d=9 " or " -18`
Hence, `c=6,b=4,d=9 " or " c=-6,b=-2,d=-18`