The number of one element subset, the number of two element subsets and the number of three element subsets of a set containing more than three elements are consecutive terms of an A.P. The number of elements in the set can be,

Question

The number of one element subset, the number of two element subsets and the number of three element subsets of a set containing more than three elements are consecutive terms of an A.P. The number of elements in the set can be, (A) 2 (B) 5 (C) 7 (D) 9

Tardigrade Admin · Accepted Answer

Given n C 1, n C 2, n C 3 longrightarrow A.P. 2 ⋅ n C2= n C1+ n C3 (2 n ( n -1)/2)= n +( n ( n -1)( n -2)/6) ⇒ n 2-9 n +14=0 ∴ n =7

Q. The number of one element subset, the number of two element subsets and the number of three element subsets of a set containing more than three elements are consecutive terms of an A.P. The number of elements in the set can be,

2

5

7

9

Given $^{n} C_{1},^{n} C_{2},^{n} C_{3} ⟶$ A.P.
$2 \cdot^{n} C_{2} =^{n} C_{1} +^{n} C_{3}$
$\frac{2 n ( n - 1 )}{2} = n + \frac{n ( n - 1 ) ( n - 2 )}{6}$
$\Rightarrow n^{2} - 9 n + 14 = 0$
$∴ n = 7$

Q. The number of one element subset, the number of two element subsets and the number of three element subsets of a set containing more than three elements are consecutive terms of an A.P. The number of elements in the set can be,

2

5

7

9

Given nC1​,nC2​,nC3​⟶ A.P. 2⋅nC2​=nC1​+nC3​ 22n(n−1)​=n+6n(n−1)(n−2)​ ⇒n2−9n+14=0 ∴n=7

Given $^{n} C_{1},^{n} C_{2},^{n} C_{3} ⟶$ A.P.
$2 \cdot^{n} C_{2} =^{n} C_{1} +^{n} C_{3}$
$\frac{2 n ( n - 1 )}{2} = n + \frac{n ( n - 1 ) ( n - 2 )}{6}$
$\Rightarrow n^{2} - 9 n + 14 = 0$
$∴ n = 7$