Here are the first four terms of a sequence - OCR - GCSE Maths - Question 14 - 2018 - Paper 1

The nth term of the sequence can be expressed in the form:

$a_n = 4n + 2$

where n starts from 1. For example:

• For n = 1, $a_1 = 4(1) + 2 = 6$
• For n = 2, $a_2 = 4(2) + 2 = 10$
• For n = 3, $a_3 = 4(3) + 2 = 14$

And so on.

To determine why 511 is not a term in the sequence, we need to ascertain if 511 can be expressed as $a_n = 4n + 2$.

Rearranging gives: $4n = 511 - 2$ $4n = 509$ $n = \frac{509}{4} = 127.25$

Since n must be an integer, 511 cannot be a term in the sequence.

To find the closest term to 511, we calculate the nearest integer value for n:

$n = \frac{511 - 2}{4} = \frac{509}{4} = 127.25$

Considering n = 127 and n = 128:

• For n = 127: $a_{127} = 4(127) + 2 = 508$
• For n = 128: $a_{128} = 4(128) + 2 = 510$

Thus, the term in the sequence that is closest to 511 is 510.