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Question

The sixth term in the expansion of [{2log (103x)}+5{2(x2)log3}]m is equal to 21. If it is known that the binomial coefficient of the 2nd,3rd and 4th terms in the expansion represents respectively the first, third and fifth terms of an A.P. (the symbol log stands for logarithm to the base 10) then sum of possible values of x is

A
1
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B
3
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C
4
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D
2
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Solution

The correct option is D 2
The coefficient of the 2nd,3rd and 4th terms in the expansion are mC1,mC2 and mC3, which are given in A.P. Hence,
2 mC2=mC1+mC3
2m(m1)2!=m+m(m1)(m2)3!
m(m29m+14)=0
m(m2)(m7)=0
m=7
(for m=0 or 2 mC3 will not exists)
Now, 6th
term in the expansion, when m=7, is
T6=7C5[2log (103x)]75×[52x2log3]5=21
7×62![2log(103x)×2(x2)log 3]=21
2log(103x)+(x2)log 3=1=20
log(103x)+(x2)log 3=0
log(103x)(3)x2=0
(103x)×3x×32=1
10×3x(3x)2=9
(3x)210×3x+9=0
(3x1)(3x9)=0
3x1=03x=1=30x=0
3x9=03x=32x=2
Hence, x=0 or 2.
Sum of possible values of x is 0+2=2

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