If the 6th, 7th and 8th terms in the expansion of $(x + a)^{n}$ are respectively 112, 7 and , $\frac{1}{4}$ , find x, a and n.

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Solution

It is given that, $T_{6} = 112, T_{7} = 7$ and $T_{8} = \frac{1}{4}$

We know that, for the expanssion $(x + a)^{n}$

$\frac{T_{r + 1}}{T_{r}} = \frac{n - r + 1}{r} . \frac{a}{x}$

$∴ \frac{T_{7}}{T_{6}} = \frac{n - 6 + 1}{6} . \frac{a}{x}$ [put r = 6]

$\Rightarrow \frac{7}{112} = \frac{n - 5}{6} . \frac{a}{x}$

$\Rightarrow \frac{a}{x} = \frac{42}{112 (n - 5)}$ . . . (i)

and $\frac{T_{8}}{T_{7}} = \frac{n - 7 + 1}{7} . \frac{a}{x}$ [put r = 7]

$\Rightarrow \frac{1}{28} = \frac{n - 6}{7} . \frac{a}{x}$

$\Rightarrow \frac{a}{x} = \frac{1}{4 (n - 6)}$ . . . (ii)

$\Rightarrow \frac{42}{112 (n - 5)} = \frac{1}{4 (n - 6)}$

[by Eqs. (i) and (ii)]

$\Rightarrow$ 4 $\times$ 42 (n - 6) = 112 (n - 5)

$\Rightarrow$ 168 n - 1008 = 112n - 560

$\Rightarrow$ 56n = 448 $\Rightarrow$ n = 8

Putting n = 8 in Eq. (ii), we get

$\Rightarrow \frac{a}{x} = \frac{1}{8} \Rightarrow x = 8 a$

Now, $T_{6} =^{8} C_{5} (x)^{3} (a)^{5}$

$\Rightarrow 112 = 56 (8 a)^{3} (a)^{5}$

$\Rightarrow 2 = 512 a^{8}$

$\Rightarrow a^{8} = \frac{1}{256}$

$\Rightarrow a = \frac{1}{2}$

and n $x = 8 \times \frac{1}{2} = 4$

Hence, n = 8, x = 4

and $a = \frac{1}{2}$

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