Show that a₁, a₂, a₃, ….. form an A.P. where a_n is defined as a_n = 3 + 4n. Also find the sum of first 15 terms.

a_n = 3 + 4n
a₁ = 3 + 4 × 1 = 3 + 4 = 7
a₂ = 3 + 4 × 2 = 3 + 8 = 11
a₃ = 3 + 4 × 3 = 3 + 12 = 15
a₄ = 3 + 4 × 4 = 3 + 16 = 19.

Since, a₄ - a₃ = a₃ - a₂ = a₂ - a₁ = 4, i.e any term - preceding term = fixed number = 4.

∴ a₁, a₂, a₃ ….. form an A.P.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow S_{15} = \dfrac{15}{2}[2 \times 7 + (15 - 1)4] \\[1em] = \dfrac{15}{2}[14 + 14 \times 4] \\[1em] = \dfrac{15}{2}[14 + 56] \\[1em] = \dfrac{15}{2} \times 70 \\[1em] = 15 \times 35 \\[1em] = 525.$

Hence, the sum of first 15 terms of the A.P. is 525.