The (r – 1)th term is

n

‌

Cr–2(x)n–r+2 & its coefficient is

n

‌

Cr–2. Similarly,the coefficients of rth & (r + 1)th terms are

n

‌

Cr–1 &

n

‌

Cr, respectively. Since the coefficients are in the ratio 1 : 3 : 5, so we have

‌nCr−2

n ‌ Cr−1

=

1

3

i.e.

r−1

n−r+2

=

1

3

i.e., n - r + 2 = 3r - 3 ⇒ n - 4r + 5 = 0 ... (i)
and

‌nCr−1

n ‌ Cr

=

3

5

i.e.,

r

n−r+1

=

3

5

i.e. 3n – 3r + 3 = 5r ⇒ 3n – 8r + 3 = 0 ... (ii)
Solving (i) & (ii), we get n = 7, r = 3.