If the coefficients of $r$ th and $(r + 1)$ th terms in the expansion of $(3 + 7 x)^{29}$ are equal, then $r$ equals?

15

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21

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14

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None of these

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Solution

The correct option is D None of these
$(a + b x)^{n} = T_{r} = n C_{r} a^{n - r} (b x)^{r} =^{n} C_{r} a^{n - r} b^{r} c o e f f i c i e n t$
$a = 3, b = 7, n = 29$
$T_{r} = 29_{C_{r}} (3)^{29 - r}$
$T_{r + 1} = 29_{C_{r + 1}} (3)^{29 - r - 1} (7)^{r + 1}$
Given, $T_{r} = T_{r + 1}$
$T_{r} = T_{r + 1} = \frac{29!}{r! (29 - r)!} \times (3)^{29 - r} 7^{r}$
$= \frac{(29)!}{(r + 1)! (29 - r - 1)!} \times (3)^{29 - r - 1} (7)^{r + 1}$
$= \frac{(3)^{29 - r - 28 + r}}{r! (29 - r) (28 - r)!} = \frac{7^{r + 1 - r}}{(r + 1) r! (28 - r)!}$
$= \frac{3}{29 - r} = \frac{7}{(r + 1)}$
$= 3 r + 3 = 203 - 7 r$
$= 10 r = 200$
$r = 20$

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