a_n = 3+4n

a₁ = 3+4(1) = 7

a₂ = 3+4(2) = 3+8 = 11

a₃ = 3+4(3) = 3+12 = 15

a₄ = 3+4(4) = 3+16 = 19

We can see here, the common difference between the terms are;

a₂ − a₁ = 11−7 = 4

a₃ − a₂ = 15−11 = 4

a₄ − a₃ = 19−15 = 4

Hence, a_{k + 1} − a_k is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of nth term is;

S_n = n/2[2a+(n -1)d]

S₁₅ = 15/2[2(7)+(15-1)×4]

= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525