The first three terms of the expansion $(a + b)^n$ are $\,,^{n}C_0 a^n , \,,^{n}C_1 a^{n-1} b , \,,^{n}C_2 a^{n-2} b^2$. Then $\,,^{n}C_0 a^n$ = 729 ⇒ $a^n$ = 729 ... (i) $\,,^{n}C_0 a^{n-1}b$ = 7290 ⇒ $na^{n-1}b$ = 7290 ... (ii) $\,,^{n}C_2 a^{n-2} b^2$ = 30375 ⇒ $\frac{n(n-1)}{2} a^{n-2} b^2$ = 30375 ... (iii) Multiplying (i) & (iii), we get $\frac{n(n-1)}{2} a^{2n-2} b^2$ = 729 × 30375 ... (iv) Squaring (ii) we get $n^2 a^{2n-2} b^2$ = 7290 × 7290 ... (v) Dividing (iv) by (v), we get $\frac{n-1}{2n}$ = $\frac{6075}{2 \times 7290}$ = $\frac{5}{12}$ ⇒ 6n - 6 = 5n ⇒ n = 6 Put n = 6 in (i) we get $a^6$ = 729 ⇒ a = 3 Substitute n = 6, a = 3 in (ii), we get $6(3)^5 b$ = 7290 ⇒ b = $\frac{7290}{6 \times (3)^5}$ ⇒ b = 5 Thus, a = 3, b = 5, n = 6.