Let the sixth term in the binomial expansion of (√2^ log_2(10-3^x)+√[5]2^(x-2) log_23)^m powers of 2^(x-2) log_23, be 21 . If the binomial…
JEE Main 2023 — Mathematics Algebra
2023integereasy
Let the sixth term in the binomial expansion of (2log2(10−3x)+52(x−2)log23)m powers of 2(x−2)log23, be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is _____ .
Official previous-year question
Held on 1 Feb 2023 · Verified 6 Jul 2026.
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Solution
Given,
Binomial expression,
(2log2(10−3x)+52(x−2)log23)m
((10−3x)+53(x−2))m
Now, T6=C5m(10−3x)2m−5⋅(3x−2)=21…(1)
Also given,
C1m,C2m,C3m are in A.P.
So, 2⋅C2m=C1m+C3m
⇒2×2!(m−2)!m!=m+3!(m−3)!m!
Solving for m, we get m=2,7 and m=2 (rejected), so m=7
Put in equation (1)
21⋅(10−3x)93x=21
⇒(10−3x)3x=9×1
⇒3x=30,32
⇒x=0,2
Sum of the squares of all possible values of x=4.
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