Find the coefficient of $x^{7}$ in $(a x^{2} + \frac{1}{b x})$ and $x^{- 7}$ in $(a x - \frac{1}{b x^{2}})$ and find the relation between a and b, so that these coefficients are equal.

Or

In the binomial expansion of $(1 + x)^{n}$ , the coefficients of the fifth, sixth and seventh terms are in AP. Find all values of n for which this can happen.

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Solution

Suppose $x^{7}$ occurs in (r+1)th term of the expansion of ${(a x^{2} + \frac{1}{b x})}^{11}$ .

Then, $T_{r + 1} = 11_{C_{r}} (a x^{2})^{11 - r} {(\frac{1}{b x})}^{r}$

$\Rightarrow T_{r + 1} = 11_{C_{6}} a^{11 - r} x^{22 - 3 r} b^{- r}$ ...(i)

On putting r=5 in Eq. (i), we get

$T_{6} = 11_{C_{5}} a^{11 - 5} b^{- 5} x^{22 - 15} = 11_{C_{5}} a^{6} b^{- 5} x^{7}$

$∴$ Coefficient of $x^{7}$ in the expansion of ${(a x^{2} + \frac{1}{b x})}^{11}$ is $11_{C_{5}} a^{6} b^{- 5}$ .

Suppose $x^{- 7}$ occurs in (r+1)th term of the expansion of ${(a x - \frac{1}{b x^{2}})}^{11}$ .

Then, $T_{r + 1} = 11_{C_{r}} (a x)^{11 - r} (- 1)^{r} {(\frac{1}{b x^{2}})}^{r}$ .

$\Rightarrow T_{r + 1} = (- 1)^{r} 11_{C_{r}} a^{11 - r} b^{- r} x^{11 - 3 r}$ ...(ii)

This will contain $x^{- 7}$ , if $11 - 3 r = - 7 \Rightarrow 3 r = 18 \Rightarrow r = 6$

On putting r=6 in Eq. (ii), we get

$T_{7} = (- 1)^{6} 11_{C_{6}} a^{11 - 6} b^{- 6} x^{11 - 3 \times 6} = 11_{C_{6}} a^{5} b^{- 6} x^{- 7}$

$∴$ Coefficient of $x^{- 7}$ in the expansion of ${(a x - \frac{1}{b x^{2}})}^{11}$ is $11_{C_{6}} a^{5} b^{- 6}$ .

If the coefficient of $x^{7}$ in ${(a x^{2} + \frac{1}{b x})}^{11}$ is equal to the coefficient of $x^{- 7}$ in ${(a x - \frac{1}{b x^{2}})}^{11}$ , then

$11_{C_{5}} a^{6} b^{- 5} = 11_{C_{6}} a^{5} b^{- 6}$

$\Rightarrow 11_{C_{5}} a b = 11_{C_{6}}$

$∴ a b = 1$

Or

The coefficients of fifth, sixth and seventh terms in the binomial expansion of $(1 + x)^{n}$ are $n_{C_{4}}$ , $n_{C_{5}}$ and $n_{C_{6}}$ , respectively.

We have, given $n_{C_{4}}, n_{C_{5}}$ and $n_{C_{6}}$ are in A.P.

$2^{n}, C_{5} = n_{C_{4}} + n_{C_{6}}$

$\Rightarrow 2 = \frac{n_{C_{4}}}{n_{C_{5}}} + \frac{n_{C_{6}}}{n_{C_{5}}}$

$\Rightarrow 2 = \frac{5}{n - 4} + \frac{n - 5}{6}$

$\Rightarrow 2 = \frac{30 + (n - 5) (n - 4)}{6 (n - 4)}$

$\Rightarrow 12 n - 48 = 30 + n^{2} - 9 n + 20$

$\Rightarrow n^{2} - 21 n + 98 = 0 \Rightarrow n^{2} - 14 n - 7 n + 98 = 0$

$\Rightarrow n (n - 14) - 7 (n - 14) = 0 \Rightarrow (n - 14) (n - 7) = 0$

$∴ n = 7, 14$

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Find the coefficient of x7 in (ax2+1bx) and x−7 in (ax−1bx2) and find the relation between a and b, so that these coefficients are equal. Or In the binomial expansion of (1+x)n, the coefficients of the fifth, sixth and seventh terms are in AP. Find all values of n for which this can happen.