Binomial Theorem — JEE Main · 2023 · 01 Feb 2023 · Shift 2 Question with Solution

Let the sixth term in the binomial expansion of

(2^{l o g_{2} (10 - 3^{x})} + 5 2^{(x - 2) l o g_{2} 3})^{m}

, in the increasing powers of

2^{(x - 2) l o g_{2} 3}

, 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 ______.

Answer: 4

Solution

The binomial coefficients

(1 m), (2 m), (3 m)

are the first, third and fifth terms of an arithmetic progression, hence themselves in arithmetic progression, so

2 (2 m) = (1 m) + (3 m)

. This gives

m^{2} - 9 m + 14 = 0

, so

m = 7

(the value

m = 2

is too small for a sixth term). Here

2^{l o g_{2} (10 - 3^{x})} = 10 - 3^{x}

and

5 2^{(x - 2) l o g_{2} 3} = 3^{(x - 2) /5}

, so the sixth term is

(5 7) (10 - 3^{x}) 3^{x - 2} = 21

, that is

(10 - 3^{x}) 3^{x} = 9

. Writing

w = 3^{x}

gives

w^{2} - 10 w + 9 = 0

, so

w = 1

w = 9

, i.e.

x = 0

x = 2

. The sum of the squares of these values is

0^{2} + 2^{2} = 4

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Binomial Theorem

Solution

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