The coefficients of rth  and (r+1)th  terms in the expansion of (1+x)20  are in the ratio 1:2,  then r=

Given expansion of (1+x)20 (∴  Tr+1= nCr xn−r (a)r  be the expansion of (x+a)n)  Tr+1= 20Cr (1)20−r xr ⇒  coefficient  of (r+1)th  term = 20Cr AndT(r−1)+1= 20Cr−1 (1)20−r+1 xr−1  Hence coefficient of rth  term = 20Cr−1  According to problem the coefficients of rth  and (r+1)th  terms in the expansion of (1+x)20  are in the ratio 1:2, thenWe get  that 20Cr−120Cr=12 ⇒ 2021−r r−1×20−r20=12  (∴ nCr=n!(n−r)! r!) ⇒r21−r=12⇒2r=21−r  ∴r=7