Question

If the coefficient of the ${{5}^{th}}$, ${{6}^{th}}$ and ${{7}^{th}}$ term is the expansion ${{\left( 1+x \right)}^{n}}$ are in AP then n =
(a). 7
(b). 5
(c). 3
(d). 10

Answer 1

- Hint: Find the terms of ${{5}^{th}}$, ${{6}^{th}}$ and ${{7}^{th}}$ then are in AP and establish their relation using basic formula of AP. Expand them in the form of ${}^{n}{{C}_{r}}$. Simplify and solve the quadratic formula thus obtained to get value of n.

Complete step-by-step solution -

It is said that the coefficient of ${{5}^{th}}$, ${{6}^{th}}$ and ${{7}^{th}}$ term is in the expansion of ${{\left( 1+x \right)}^{n}}$. The binomial expansion of ${{\left( 1+x \right)}^{n}}$ is of the form,
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+.....+{}^{n}{{C}_{n}}{{x}^{n}}-(1)$
It is also said that ${{5}^{th}}$, ${{6}^{th}}$ and ${{7}^{th}}$ term are in AP. The ${{n}^{th}}$ term of an AP is represented as ${{T}_{n}}$.
$\therefore$ ${{5}^{th}}$ term of an AP is ${{T}_{5}}$. Similarly, ${{6}^{th}}$ and ${{7}^{th}}$ term of an AP can be represented as ${{T}_{6}}$ and ${{T}_{5}}$.
Thus we can say that ${{T}_{5}}$, ${{T}_{6}}$ and ${{T}_{7}}$ are in AP.
We know that AP represents arithmetic progression. It is a sequence of numbers such that the difference between the consecutive terms is constant.
Difference here means ${{2}^{nd}}$ minus the first term. Common difference is denoted as ‘d’.
Here, ${{T}_{5}}$, ${{T}_{6}}$ and ${{T}_{7}}$ are in AP.
$\therefore$ Common difference, $d={{T}_{6}}-{{T}_{5}}$.
Similarly, $d={{T}_{7}}-{{T}_{6}}$.
Thus from the above we can write,

& {{T}_{6}}-{{T}_{5}}={{T}_{7}}-{{T}_{6}} \\\ & \Rightarrow {{T}_{6}}+{{T}_{6}}={{T}_{5}}+{{T}_{7}} \\\ & 2{{T}_{6}}={{T}_{5}}+{{T}_{7}}-(2) \\\ \end{aligned}$$ We can write, $${{T}_{6}}={}^{n}{{C}_{5}},{{T}_{5}}={}^{n}{{C}_{4}}$$ and $${{T}_{7}}={}^{n}{{C}_{6}}$$. Thus equation (2) becomes, $$2{}^{n}{{C}_{5}}={}^{n}{{C}_{4}}+{}^{n}{{C}_{6}}$$ It is of the form, $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$. Thus we can expand the above expression as, $$\begin{aligned} & \dfrac{2n!}{5!\left( n-5 \right)!}=\dfrac{n!}{4!\left( n-4 \right)!}+\dfrac{n!}{6!\left( n-6 \right)!} \\\ & \dfrac{2n!}{\left( n-5 \right)\left( n-6 \right)!5\times 4!}=\dfrac{n!}{\left( n-4 \right)\left( n-5 \right)\left( n-6 \right)!4!}+\dfrac{n!}{\left( n-6 \right)!6\times 5\times 4!} \\\ & \dfrac{n!}{4!\left( n-6 \right)!}\left[ \dfrac{2}{\left( n-5 \right)\times 5} \right]=\dfrac{n!}{\left( n-6 \right)!4!}\left[ \dfrac{1}{\left( n-6 \right)\left( n-5 \right)}+\dfrac{1}{6\times 5} \right] \\\ \end{aligned}$$ Cancel out $$\dfrac{n!}{\left( n-6 \right)!4!}$$ from both sides of the expression and simplify it. $$\begin{aligned} & \dfrac{2}{5\left( n-5 \right)}=\dfrac{1}{\left( n-4 \right)\left( n-5 \right)}+\dfrac{1}{6\times 5} \\\ & \dfrac{2}{\left( n-5 \right)}=5\left[ \dfrac{1}{\left( n-4 \right)\left( n-5 \right)}+\dfrac{1}{6\times 5} \right] \\\ \end{aligned}$$ $$\Rightarrow \dfrac{2}{n-5}=\dfrac{5}{\left( n-4 \right)\left( n-5 \right)}+\dfrac{1}{6}$$, Rearrange the equation and simplify it. $$\begin{aligned} & \dfrac{2}{n-5}-\dfrac{5}{\left( n-4 \right)\left( n-5 \right)}=\dfrac{1}{6} \\\ & \dfrac{1}{\left( n-5 \right)}\left[ 2-\dfrac{5}{n-4} \right]=\dfrac{1}{6} \\\ \end{aligned}$$ $$\dfrac{2\left( n-4 \right)-5}{\left( n-4 \right)\left( n-5 \right)}=\dfrac{1}{6}$$, Cross multiply the expression. $$\begin{aligned} & 6\left[ 2n-8-5 \right]=\left( n-4 \right)\left( n-5 \right) \\\ & 6\left[ 2n-13 \right]={{n}^{2}}-4n-5n+20 \\\ & 12n-78={{n}^{2}}-9n+20 \\\ & \Rightarrow {{n}^{2}}-21n+98=0-(3) \\\ \end{aligned}$$ The equation (3) is similar to the general quadratic equation, $$a{{x}^{2}}+bx+c=0$$. It is of the form $${{n}^{2}}$$ - (Sum of zeroes) n + (Product of zeroes) = 0. Let us consider 2 terms as a and b. i.e. Sum of zeroes = a + b = -21. Product of zeroes = ab = 98 Now let us find a and b. Try putting, a = -7 and b = -14. Sum of zeroes = a + b = (-7) + (-14) = -21. Product of zeroes = ab = (-7) (-14) = 98 Thus the values are the same. Now split the $${{2}^{nd}}$$ term of equation (3) as (-7n – 14n). i.e. $${{n}^{2}}-21n+98=0$$ $$\begin{aligned} & {{n}^{2}}-7n-14n+98=0 \\\ & n\left( n-7 \right)-14\left( n-7 \right)=0 \\\ & \left( n-7 \right)\left( n-14 \right)=0 \\\ \end{aligned}$$ i.e. n – 7 = 0 and n – 14 = 0 Thus we got n as 7 and 14. From the options given, n = 7 is the correct answer. Thus we got, n = 7. $$\therefore $$ Option (a) is the correct answer. Note: n =14 is also the correct answer. But checking the options provided there is number n = 14. So we can neglect that value and choose the correct answer as n = 7. Remember the basic formulas of arithmetic progression and that of $${}^{n}{{C}_{r}}$$ to solve problems like this.