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,
\[\begin{align}
  & {{T}_{6}}-{{T}_{5}}={{T}_{7}}-{{T}_{6}} \\
 & \Rightarrow {{T}_{6}}+{{T}_{6}}={{T}_{5}}+{{T}_{7}} \\
 & 2{{T}_{6}}={{T}_{5}}+{{T}_{7}}-(2) \\
\end{align}\]
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,
\[\begin{align}
  & \dfrac{2n!}{5!\left( n-5 \right)!}=\dfrac{n!}{4!\left( n-4 \right)!}+\dfrac{n!}{6!\left( n-6 \right)!} \\
 & \dfrac{2n!}{\left( n-5 \right)\left( n-6 \right)!5\times 4!}=\dfrac{n!}{\left( n-4 \right)\left( n-5 \right)\left( n-6 \right)!4!}+\dfrac{n!}{\left( n-6 \right)!6\times 5\times 4!} \\
 & \dfrac{n!}{4!\left( n-6 \right)!}\left[ \dfrac{2}{\left( n-5 \right)\times 5} \right]=\dfrac{n!}{\left( n-6 \right)!4!}\left[ \dfrac{1}{\left( n-6 \right)\left( n-5 \right)}+\dfrac{1}{6\times 5} \right] \\
\end{align}\]
Cancel out \[\dfrac{n!}{\left( n-6 \right)!4!}\] from both sides of the expression and simplify it.
\[\begin{align}
  & \dfrac{2}{5\left( n-5 \right)}=\dfrac{1}{\left( n-4 \right)\left( n-5 \right)}+\dfrac{1}{6\times 5} \\
 & \dfrac{2}{\left( n-5 \right)}=5\left[ \dfrac{1}{\left( n-4 \right)\left( n-5 \right)}+\dfrac{1}{6\times 5} \right] \\
\end{align}\]
\[\Rightarrow \dfrac{2}{n-5}=\dfrac{5}{\left( n-4 \right)\left( n-5 \right)}+\dfrac{1}{6}\], Rearrange the equation and simplify it.
\[\begin{align}
  & \dfrac{2}{n-5}-\dfrac{5}{\left( n-4 \right)\left( n-5 \right)}=\dfrac{1}{6} \\
 & \dfrac{1}{\left( n-5 \right)}\left[ 2-\dfrac{5}{n-4} \right]=\dfrac{1}{6} \\
\end{align}\]
\[\dfrac{2\left( n-4 \right)-5}{\left( n-4 \right)\left( n-5 \right)}=\dfrac{1}{6}\], Cross multiply the expression.
\[\begin{align}
  & 6\left[ 2n-8-5 \right]=\left( n-4 \right)\left( n-5 \right) \\
 & 6\left[ 2n-13 \right]={{n}^{2}}-4n-5n+20 \\
 & 12n-78={{n}^{2}}-9n+20 \\
 & \Rightarrow {{n}^{2}}-21n+98=0-(3) \\
\end{align}\]
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. \[{{n}^{2}}-21n+98=0\]
\[\begin{align}
  & {{n}^{2}}-7n-14n+98=0 \\
 & n\left( n-7 \right)-14\left( n-7 \right)=0 \\
 & \left( n-7 \right)\left( n-14 \right)=0 \\
\end{align}\]
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.