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Relation between Coefficient and Indices of x and y

Find the two ...

Question

Find the two successive terms in the expansion of $(3 + 4 x)^{7}$ whose coefficients of x are in the ratio 1:1.

T_{5}

T_{6}

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T_{3}

T_{4}

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T_{4}

T_{5}

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Does not exist

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Solution

The correct option is D Does not exist
Given, $(3 + 4 x)^{7}$

General term $T_{r + 1} =^{n} C_{r} x^{n - r} y^{r}$
$T_{r + 1} =^{7} C_{r} (3)^{7 - r} (4 x)^{r}$
$T_{r + 1} =^{7} C_{r} (3)^{7 - r} (4)^{r} x^{r}$
$T_{r + 2} = T_{2 + 1 + 1} =^{7} C_{r + 1} (3)^{6 - r} (4)^{r + 1} x^{r + 1}$

Given coefficient of x are in ratio
$\frac{^{7} C_{r} (3)^{7 - r} 4^{r}}{^{7} C_{r + 1} (3)^{6 - r} (4)^{r + 1}}$
$\frac{7! (r + 1)! (6 - r)!}{r! (7 - r)! \times 7!} \frac{(3)^{1}}{(4)^{1}}$
$\frac{(r + 1) (r)! \times (6 - r)! 3}{r! (7 - r) (6 - r)! \times 4}$
$= \frac{3 (r + 1)}{4 (7 - r)} = 1$
$3 r + 3 = 28 - 4 r$
$7 r = 25$
$r = \frac{25}{7}$
Hence does not exist because $r$ cannot be fraction.

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