Let the coefficients of three consecutive terms $\mathrm{T}_{\mathrm{r}}$, $\mathrm{T}_{\mathrm{r}+1}$ and $\mathrm{T}_{\mathrm{r}+2}$ in the binomial expansion of $(\mathrm{a}+\mathrm{b})^{12}$ be in a G.P. and let p be the number of all possible values of r . Let q be the sum of all rational terms in the binomial expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$. Then $\mathrm{p}+\mathrm{q}$ is equal to :

Binomial Theorem — JEE Main · 2025 · 28 Jan 2025 · Shift 2 Question with Solution

Let the coefficients of three consecutive terms

T_{r}

T_{r + 1}

and

T_{r + 2}

in the binomial expansion of

(a + b)^{12}

be in a G.P. and let p be the number of all possible values of r . Let q be the sum of all rational terms in the binomial expansion of

(43 + 34)^{12}

. Then

p + q

is equal to :

Answer: (A)

For

(a + b)^{12}

, write

T_{r} = (r - 1 12)

T_{r + 1} = (r 12)

T_{r + 2} = (r + 1 12)

. These are in G.P. when

(r 12)^{2} = (r - 1 12) (r + 1 12)

Using

\frac{( r 12 )}{( r - 1 12 )} = \frac{13 - r}{r}

and

\frac{( r + 1 12 )}{( r 12 )} = \frac{12 - r}{r + 1}

, the G.P. condition becomes

\frac{13 - r}{r} = \frac{12 - r}{r + 1}

(13 - r) (r + 1) = r (12 - r) ⟹ - r^{2} + 12 r + 13 = - r^{2} + 12 r ⟹ 13 = 0

This is never true, so no value of

r

works and

p = 0

.

For

q

, expand

(3^{1/4} + 4^{1/3})^{12} = \sum_{r = 0}^{12} (r 12) 3^{(12 - r) /4} 4^{r /3}

. A term is rational when both

\frac{12 - r}{4}

and

\frac{r}{3}

are integers, which forces

r

to be a multiple of

lcm (4, 3) = 12

; so

r = 0

r = 12

r = 0

(0 12) 3^{3} = 27

r = 12

(12 12) 4^{4} = 256

q = 27 + 256 = 283

, and

p + q = 0 + 283 = 283

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option