If the coefficients of rth, (r+ 1)th and (r + 2)th terms in
the expansion of `(1 + x)^(1//4)` are in AP, then r is /are

Options

A. 5
B. 9
C. 10
D. 12

Answer

a,b

Explanation

Coefficient of rth (r+1) th and (r +2) th terms in ` ( 1 + x)^(14)` are
` ""^(14)C_(r-1), ""^(14)C_(r)`
and ` ""^(14)C_(r-1)`,respectively
Now , according to the question , ` 2 (""^(14)C_(r)) = ""^(14)C_(r-1) + ""^(14)C_(r+1)`
On dividing both sides by ` ""^(14)C_(r-1) ` , we get
`2= (""^(14)C_(r-1))/(""^(14)C_(r)) = (""^(14)C_(r+1))/(""^(14)C_(r))`
` rArr 2= (r)/(14 - r + 1) + (14 - (r +1)+1)/(r +1)`
` rArr 2 = (r)/(15-r) + (14-r)/(r+1)`
` rArr 2(15-r) (r+1)= r(r+1) + (15 - r) (14 - r)`
` rArr - 2r^(2) + 28r + 30 + 2r^(2) - 28 r + 210`
` rArr 4t^(2) - 56r + 180 = 0 rArr r^(2) - 14r + 45 = 0 `
` rArr (r-9)(r-5) = 0 `
` rArr r = 5,9`