if the 6th,7th and 8th terms in the expansion of (x+a)ⁿ are respectively 112,7 and 1/4,find x, a and n.

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\left(\mathrm{x}+\mathrm{a}\right)}^{\mathrm{n}} \\ {\mathrm{T}}_6={\mathrm{T}}_{5+1}{=}^{\mathrm{n}}{\mathrm{C}}_5 {\mathrm{x}}^{\mathrm{n}-5}{\mathrm{a}}^5=112...\left(\mathrm{i}\right) \\ {\mathrm{T}}_7={\mathrm{T}}_{6+1}{=}^{\mathrm{n}}{\mathrm{C}}_6 {\mathrm{x}}^{\mathrm{n}-6}{\mathrm{a}}^6=7...\left(\mathrm{ii}\right) \\ {\mathrm{T}}_8={\mathrm{T}}_{7+1}{=}^{\mathrm{n}}{\mathrm{C}}_7 {\mathrm{x}}^{\mathrm{n}-7}{\mathrm{a}}^7=\frac{1}{4}...\left(\mathrm{iii}\right) \\ \frac{\left(\mathrm{i}\right)}{\left(\mathrm{ii}\right)} \\ \frac{{}^{\mathrm{n}}{\mathrm{C}}_5 {\mathrm{x}}^{\mathrm{n}-5}{\mathrm{a}}^5}{{}^{\mathrm{n}}{\mathrm{C}}_6 {\mathrm{x}}^{\mathrm{n}-6}{\mathrm{a}}^6}=\frac{112}{7} \\ \Rightarrow \frac{{\displaystyle \frac{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)\left(\mathrm{n}-3\right)\left(\mathrm{n}-4\right)}{5!}}}{\frac{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)\left(\mathrm{n}-3\right)\left(\mathrm{n}-4\right)\left(\mathrm{n}-5\right)}{6.5!}}.\frac{\mathrm{x}}{\mathrm{a}}=16 \\ \Rightarrow \frac{6}{\mathrm{n}-5}.\frac{\mathrm{x}}{\mathrm{a}}=16...\left(\mathrm{iv}\right) \\ \frac{\left(\mathrm{ii}\right)}{\left(\mathrm{iii}\right)} \\ \frac{{}^{\mathrm{n}}{\mathrm{C}}_6 {\mathrm{x}}^{\mathrm{n}-6}{\mathrm{a}}^6}{{}^{\mathrm{n}}{\mathrm{C}}_7 {\mathrm{x}}^{\mathrm{n}-7}{\mathrm{a}}^7}=\frac{7}{{\displaystyle \frac{1}{4}}} \\ \Rightarrow \frac{\frac{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)\left(\mathrm{n}-3\right)\left(\mathrm{n}-4\right)\left(\mathrm{n}-5\right)}{6!}}{\frac{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)\left(\mathrm{n}-3\right)\left(\mathrm{n}-4\right)\left(\mathrm{n}-5\right)\left(\mathrm{n}-6\right)}{7.6!}}.\frac{\mathrm{x}}{\mathrm{a}}=28 \\ \Rightarrow \frac{7}{\mathrm{n}-6}.\frac{\mathrm{x}}{\mathrm{a}}=28...\left(\mathrm{v}\right) \\ \left(\mathrm{v}\right)/\left(\mathrm{iv}\right) \\ \frac{\frac{7}{\mathrm{n}-6}}{\frac{6}{\mathrm{n}-5}}=\frac{28}{16}=\frac{7}{4} \\ \Rightarrow 28\left(\mathrm{n}-5\right)=42\left(\mathrm{n}-6\right) \\ \Rightarrow 4\left(\mathrm{n}-5\right)=6\left(\mathrm{n}-6\right) \\ \Rightarrow 4\mathrm{n}-20=6\mathrm{n}-36 \\ \Rightarrow 2\mathrm{n}=16 \\ \Rightarrow \mathrm{n}=8 \\ \mathrm{Hence} \left(\mathrm{v}\right) \mathrm{becomes} \\ \frac{7}{2}.\frac{\mathrm{x}}{\mathrm{a}}=28 \\ \Rightarrow \mathrm{x}=8\mathrm{a} \\ \mathrm{Put} \mathrm{this} \mathrm{in} \left(\mathrm{i}\right){ \\ }^{\mathrm{n}}{\mathrm{C}}_5 {\mathrm{x}}^{\mathrm{n}-5}{\mathrm{a}}^5=112{ \\ }^8{\mathrm{C}}_5 {\mathrm{x}}^{8-5}{\mathrm{a}}^5=112 \\ {\Rightarrow }^8{\mathrm{C}}_5 {\mathrm{x}}^3{\mathrm{a}}^5=112 \\ {\Rightarrow }^8{\mathrm{C}}_5 {\left(8\mathrm{a}\right)}^3{\mathrm{a}}^5=112 \\ \Rightarrow {\mathrm{a}}^8=\frac{112}{8^3{.}^8{\mathrm{C}}_5} \\ =\frac{112}{8^3.{\displaystyle \frac{8\times 7\times 6}{3\times 2\times 1}}} \\ =\frac{16}{8^4} \\ =\frac{2^4}{2^{12}} \\ \Rightarrow {\mathrm{a}}^8={\left(\frac{1}{2}\right)}^8 \\ \Rightarrow \mathrm{a}=\frac{1}{2} \\ \mathrm{x}=8\mathrm{a}=4 \\ \mathrm{Hence} \\ \mathrm{a}=\frac{1}{2};\mathrm{x}=4; \mathrm{n}=8 \end{gathered}$$

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