The (r – 1)th term is $\,{}^{n}C_{r-2} (x)^{n-r+2}$ & its coefficient is $\,{}^{n}C_{r-2}$. Similarly,the coefficients of rth & (r + 1)th terms are $\,{}^{n}C_{r-1}$ & $\,{}^{n}C_{r}$, respectively. Since the coefficients are in the ratio 1 : 3 : 5, so we have $\frac{\,nC_{r-2}}{\,{}^{n}C_{r-1}}$ = $\frac{1}{3}$ i.e. $\frac{r-1}{n-r+2}$ = $\frac{1}{3}$ i.e., n - r + 2 = 3r - 3 ⇒ n - 4r + 5 = 0 ... (i) and $\frac{\,nC_{r-1}}{\,{}^{n}C_{r}}$ = $\frac{3}{5}$ i.e., $\frac{r}{n-r+1}$ = $\frac{3}{5}$ i.e. 3n – 3r + 3 = 5r ⇒ 3n – 8r + 3 = 0 ... (ii) Solving (i) & (ii), we get n = 7, r = 3.