Consider the sequence
$8A + 2B$, $6A + B$, $4A$, $2A - B$, …… Which term of this sequence will have a coefficient of A which is twice the coefficient of B?
A. 10th
B. 14th
C. 16th
D. None of these

Hint: Here, the sequence starts with a particular first term and then to get successive terms we just add $( - 2A - B)$ to the previous term. So, the difference between two consecutive terms remains constant. Such sequences are called Arithmetic Progression (AP). If we represent the first term of the sequence by $a$ and let $d$ denote the common difference between successive terms. So, the Arithmetic Progression can be written as
$a$ , $a + d$ , $a + 2d$ , ……
Hence if we wanted to write down the nth term for the sequence, we can write as
$a + (n - 1)d$ .

Complete step by step solution: 
As provided, we have sequence $8A + 2B$, $6A + B$, $4A$, $2A - B$, …… which is an AP with common difference of $( - 2A - B)$
The first term is $a = 8A + 2B$
Now, we will find the nth term of this sequence by the formula
$a + (n - 1)d$$ = $$(8A + 2B) + (n - 1)( - 2A - B)$
                       $ = (8A + 2B) - (2An + Bn) + (2A + B)$
On further solving, we get
                       $ = (10 - 2n)A + B(3 - n)$
As per question, if we equalize the coefficient of $A$ to twice the coefficient of $B$ , we have
$10 - 2n = 2(3 - n)$
$10 - 2n = 6 - 2n$
which implies that $10 = 6$ , which is not possible
Hence, we conclude that there exists no such value of n for which the coefficient of $A$ becomes twice the coefficient of $B$ .
Therefore, the correct option is D.

Note: Increasing, decreasing, multiplying and dividing each term of an AP by a non-zero constant gives out an AP. An AP can have zero, positive or negative common difference. It is to be noted that the sum of two terms of an arithmetic progression equidistant from the beginning and end is constant and equal to the sum of first and last terms.