The first three terms of the expansion (a+b)n are

n

‌

C0an,

n

‌

C1an−1b,

n

‌

C2an−2b2. Then

n

‌

C0an = 729 ⇒ an = 729 ... (i)

n

‌

C0an−1b = 7290 ⇒ nan−1b = 7290 ... (ii)

n

‌

C2an−2b2 = 30375
⇒

n(n−1)

2

an−2b2 = 30375 ... (iii)
Multiplying (i) & (iii), we get

n(n−1)

2

a2n−2b2 = 729 × 30375 ... (iv)
Squaring (ii) we get n2a2n−2b2 = 7290 × 7290 ... (v)
Dividing (iv) by (v), we get

n−1

2n

=

6075

2×7290

=

5

12

⇒ 6n - 6 = 5n ⇒ n = 6
Put n = 6 in (i) we get a6 = 729 ⇒ a = 3
Substitute n = 6, a = 3 in (ii), we get
6(3)5b = 7290 ⇒ b =

7290

6×(3)5

⇒ b = 5
Thus, a = 3, b = 5, n = 6.