Let a=2ln(10−3x) and b=2(x−2)ln3
Then, Tr+1=mCr⋅(a)(m−r)/2⋅(b)r/5
∴T6= mC5⋅(a)(m−5)/2⋅(b)=21 ⋯(1)
Also, mC1,mC2,mC3 are in A.P.
∴2⋅mC2=mC1+mC3
⇒2⋅m(m−1)2=m+m(m−1)(m−2)6
⇒6(m−1)=6+(m−1)(m−2) (∵m≠0)
⇒m=7
Substituting m=7 in equation (1), we get
7C5⋅a⋅b=21
⇒ab=1
⇒2ln(10−3x)⋅2(x−2)ln3=1
⇒2ln(10−3x)+(x−2)ln3=20
⇒ln(10−3x)+(x−2)ln3=0
⇒ln{(10−3x)(3(x−2))}=0
⇒(10−3x)3x9=1
⇒32x−10⋅3x+9=0
Let 3x=t
∴t2−10t+9=0
⇒t=1 or t=9
⇒x=0 or x=2
Hence, sum of possible values of x is 0+2=2