Let a be the first term and d be the common difference.

Given: a₅ = 16

a₁₃ = 4 a₃

Now, Consider a₅ = 16

⇒ a + (5 - 1)d = 16

⇒ a + 4d = 16 ………………….(1)

Consider a₁₃ = 4 a₃

⇒ a + 12d = 4(a + 2d)

⇒ a + 12d = 4a + 8d

⇒ 3a - 4d = 0 ………………….(2)

Now, adding equation (1) and (2), we get,

4a = 16

⇒ a = 4

∴ from equation (2), we get,

4d = 3a

⇒ 4d = 12

⇒ d = 3

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 10 terms is given by:

S₁₀ = [2(4) + (10 - 1)(3)]

= 5 × [8 + 27]

= 5 × 35

= 175

∴ S₁₀=175