Question

If $^n{C_4}{,^n}{C_5}$ and $^n{C_6}$ are in A.P then $n$ is
(a) 7 or 14
(b) 7
(c) 14
(d) 14 or 21

Answer 1

Hint: We will first expand the given terms using $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Since, the given terms are in A.P. then, the difference of first and second term is equal to the difference of second and third term. We will substitute the required values and then solve for the value of $n$

Complete step-by-step answer:
We will first write the values of $^n{C_4}{,^n}{C_5}$ and $^n{C_6}$.
As we know that $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Then,
\[^n{C_4} = \dfrac{{n!}}{{4!\left( {n - 4} \right)!}}\],
\[^n{C_5} = \dfrac{{n!}}{{5!\left( {n - 5} \right)!}}\] and
\[^n{C_6} = \dfrac{{n!}}{{6!\left( {n - 6} \right)!}}\]
We are given that $^n{C_4}{,^n}{C_5}$ and $^n{C_6}$ are in A.P, that they have equal common difference.
\[\dfrac{{n!}}{{5!\left( {n - 5} \right)!}} - \dfrac{{n!}}{{4!\left( {n - 4} \right)!}} = \dfrac{{n!}}{{6!\left( {n - 6} \right)!}} - \dfrac{{n!}}{{5!\left( {n - 5} \right)!}}\]
We will take $n!$ common and cancel from both sides.
Then, simplify the terms by taking common terms out.
$
  \dfrac{1}{{5!\left( {n - 5} \right)!}} - \dfrac{1}{{4!\left( {n - 4} \right)!}} = \dfrac{1}{{6!\left( {n - 6} \right)!}} - \dfrac{1}{{5!\left( {n - 5} \right)!}} \\
   \Rightarrow \dfrac{1}{{4!\left( {n - 5} \right)!}}\left( {\dfrac{1}{5} - \dfrac{1}{{\left( {n - 4} \right)}}} \right) = \dfrac{1}{{5!\left( {n - 6} \right)!}}\left( {\dfrac{1}{6} - \dfrac{1}{{\left( {n - 5} \right)}}} \right) \\
$
Since, we know $n! = n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1$
$
  \dfrac{1}{{4!\left( {n - 5} \right).\left( {n - 6} \right)!}}\left( {\dfrac{1}{5} - \dfrac{1}{{\left( {n - 4} \right)}}} \right) = \dfrac{1}{{5.4!\left( {n - 6} \right)!}}\left( {\dfrac{1}{6} - \dfrac{1}{{\left( {n - 5} \right)}}} \right) \\
   \Rightarrow \dfrac{1}{{n - 5}}\left( {\dfrac{{n - 4 - 5}}{{5\left( {n - 4} \right)}}} \right) = \dfrac{1}{5}\left( {\dfrac{{n - 5 - 6}}{{6\left( {n - 5} \right)}}} \right) \\
   \Rightarrow 6\left( {\dfrac{{n - 9}}{{n - 4}}} \right) = \left( {n - 11} \right) \\
$
Then, we will simplify the equation,
\[6\left( {\dfrac{{n - 9}}{{n - 4}}} \right) - \left( {n - 11} \right) = 0\]
Take LCM and then solve the for the value of $n$
$
  \left( {\dfrac{{6\left( {n - 9} \right) - \left( {n - 11} \right)\left( {n - 4} \right)}}{{n - 4}}} \right) = 0 \\
   \Rightarrow 6\left( {n - 9} \right) - \left( {n - 11} \right)\left( {n - 4} \right) = 0 \\
   \Rightarrow 6n - 54 - {n^2} + 15n - 44 = 0 \\
   \Rightarrow - {n^2} + 21n - 98 = 0 \\
   \Rightarrow {n^2} - 21n + 98 = 0 \\
$
Factorise the above equation,
$
  {n^2} - 21n + 98 = 0 \\
   \Rightarrow {n^2} - 14n - 7n + 98 = 0 \\
   \Rightarrow n\left( {n - 14} \right) - 7\left( {n - 14} \right) = 0 \\
   \Rightarrow \left( {n - 7} \right)\left( {n - 14} \right) = 0 \\
$
Equate the factors to 0 to find the value of $n$
$
  \left( {n - 7} \right) = 0 \\
  n = 7 \\
$
And
$
  n - 14 = 0 \\
  n = 14 \\
$
Therefore, the value of $n$ is 7 or 14.
Hence, option A is correct.

Note: The terms that are in A.P have a common difference between them which is the same for any two consecutive terms. If a sequence ${a_1},{a_2},{a_3}....$ is an A.P., then ${a_2} - {a_1} = {a_3} - {a_2}$ or we can also form the equation, such as $2{a_2} = {a_1} + {a_3}$.