Solve for x : 1 + 4 + 7 + 10 + … + x = 287.

The above series is in A.P. because,
any term - preceding term = 3 = common difference.

Let there be n terms so, x = a_n.

Given, a = 1, d = 3 and S_n = 287.

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

$\Rightarrow S_n = \dfrac{n}{2}[2 \times 1 + (n - 1)3] \\[1em] \Rightarrow 287 = \dfrac{n}{2}[2 + 3n - 3] \\[1em] \Rightarrow 287 \times 2 = n[3n - 1] \\[1em] \Rightarrow 3n^2 - n = 574 \\[1em] \Rightarrow 3n^2 - n - 574 = 0 \\[1em] 3n^2 - 42n + 41n - 574 = 0 \\[1em] 3n(n - 14) + 41(n - 14) = 0 \\[1em] (3n + 41)(n - 14) = 0 \\[1em] 3n + 41 = 0 \text{ or } n - 14 = 0 \\[1em] n = -\dfrac{41}{3} \text{ or } n = 14.$

Since number of terms cannot be negative so n ≠ $-\dfrac{41}{3}$

∴ n = 14.

We know x = a_n so,

⇒ x = a + (n - 1)d
⇒ x = 1 + (14 - 1)3
⇒ x = 1 + 13(3)
⇒ x = 1 + 39
⇒ x = 40.

Hence, the value of x = 40.