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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 (...

If nC4,nC5^n{C_4}{,^n}{C_5} and nC6^n{C_6} are in A.P then nn is
(a) 7 or 14
(b) 7
(c) 14
(d) 14 or 21

Explanation

Solution

We will first expand the given terms using nCr=n!r!(n−r)!^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 nn

Complete step-by-step answer:
We will first write the values of nC4,nC5^n{C_4}{,^n}{C_5} and nC6^n{C_6}.
As we know that nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
Then,
nC4=n!4!(n−4)!^n{C_4} = \dfrac{{n!}}{{4!\left( {n - 4} \right)!}},
nC5=n!5!(n−5)!^n{C_5} = \dfrac{{n!}}{{5!\left( {n - 5} \right)!}} and
nC6=n!6!(n−6)!^n{C_6} = \dfrac{{n!}}{{6!\left( {n - 6} \right)!}}
We are given that nC4,nC5^n{C_4}{,^n}{C_5} and nC6^n{C_6} are in A.P, that they have equal common difference.
n!5!(n−5)!−n!4!(n−4)!=n!6!(n−6)!−n!5!(n−5)!\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!n! common and cancel from both sides.
Then, simplify the terms by taking common terms out.
15!(n−5)!−14!(n−4)!=16!(n−6)!−15!(n−5)! ⇒14!(n−5)!(15−1(n−4))=15!(n−6)!(16−1(n−5))  \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.(n−1).(n−2).....3.2.1n! = n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1
14!(n−5).(n−6)!(15−1(n−4))=15.4!(n−6)!(16−1(n−5)) ⇒1n−5(n−4−55(n−4))=15(n−5−66(n−5)) ⇒6(n−9n−4)=(n−11)  \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(n−9n−4)−(n−11)=06\left( {\dfrac{{n - 9}}{{n - 4}}} \right) - \left( {n - 11} \right) = 0
Take LCM and then solve the for the value of nn
(6(n−9)−(n−11)(n−4)n−4)=0 ⇒6(n−9)−(n−11)(n−4)=0 ⇒6n−54−n2+15n−44=0 ⇒−n2+21n−98=0 ⇒n2−21n+98=0  \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,
n2−21n+98=0 ⇒n2−14n−7n+98=0 ⇒n(n−14)−7(n−14)=0 ⇒(n−7)(n−14)=0  {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 nn
(n−7)=0 n=7  \left( {n - 7} \right) = 0 \\\ n = 7 \\\
And
n−14=0 n=14  n - 14 = 0 \\\ n = 14 \\\
Therefore, the value of nn 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 a1,a2,a3....{a_1},{a_2},{a_3}.... is an A.P., then a2−a1=a3−a2{a_2} - {a_1} = {a_3} - {a_2} or we can also form the equation, such as 2a2=a1+a32{a_2} = {a_1} + {a_3}.