Here are the first four terms of a sequence - OCR - GCSE Maths - Question 15 - 2018 - Paper 1

Based on the observed pattern, we can represent the nth term of the sequence as:

$a_n = 6 + (n - 1) imes 4$

This can be simplified to:

$a_n = 4n + 2$

where n is the position in the sequence (1 for the first term, 2 for the second term, etc.).

To determine whether 511 is a term in the sequence, we can use the expression derived in part (b):

$a_n = 4n + 2$

Setting this equal to 511, we have:

$4n + 2 = 511$

Subtracting 2 from both sides:

$4n = 509$

Dividing by 4:

n = rac{509}{4} = 127.25

Since n must be a whole number, 511 cannot be a term in the sequence.

To find the term closest to 511, we can evaluate the sequence terms around 511 using the expression for the nth term. For integer values of n:

• For n = 127: $a_{127} = 4 imes 127 + 2 = 510$

• For n = 128: $a_{128} = 4 imes 128 + 2 = 514$

Comparing these values, we see:

• The term for n = 127 is 510, which is 1 away from 511.
• The term for n = 128 is 514, which is 3 away from 511.

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