Question

If 1 + 4 + 7 + 10 + ... + x = 287, find the value of x.

Answer 1

Given: 1 + 4 + 7 + 10 + ... + x = 287, where the sequence is arithmetic with first term a = 1 and common difference d = 3.

Step-wise calculation:

1. Let n be the number of terms.

The last term x = a + (n – 1)d

= 1 + 3(n – 1)

= 3n – 2

2. Sum formula: `S = n/2 xx ("first" + "last")` 

⇒ `287 = n/2 xx (1 + x)`

3. Substitute x = 3n – 2:

`287 = n/2 xx (1 + 3n - 2)`

= `n/2 xx (3n - 1)`

4. Multiply both sides by 2:

574 = n(3n – 1)

⇒ 3n² – n – 574 = 0

5. Solve quadratic:

Discriminant Δ = (–1)² – 4 × 3 × (–574) 

= 1 + 6888

= 6889

= 83²

6. n = (1 ± 83)/(6).

Positive root: `n = (1 + 83)/6`

= `84/6`

= 14

7. x = 3n – 2

= 3 × 14 – 2 

= 42 – 2

= 40