Question

Given A.P. is $\dfrac{1}{{3q}},\dfrac{{1 - 6q}}{{3q}},\dfrac{{1 - 12q}}{{3q}}$. Find the common difference.
A) 1
B) -1
C) 2
D) -2

Answer 1

Hint: We will use the general terminology for first consecutive 3 terms for general A.P. and then establish some relation between those using any linear combination. Hence, then by putting the given terms, we will have the required answer.

Complete step-by-step answer:
Let us first get to know what an A.P. is:-
An arithmetic progression (A.P.) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
We know that if an A.P. has 3 consecutive terms.
We can write those terms as a, a + d, a + 2d.
Hence, first term = a, second term = a + d and third term = a + 2d.
We will now find: second term – first term = (a + d) – a = a + d – a = d ……………(1)
Now, let us find: third term – second term = (a + 2d) – (a + d) = a + 2d – a – d = d ……………..(2)
Equation (1) and (2), we will have:-
Common difference = d = second term – first term = third term – second term ………..(3)
Now applying this to given terms that is first term = $\dfrac{1}{{3q}}$, second term = $\dfrac{{1 - 6q}}{{3q}}$ and third term = $\dfrac{{1 - 12q}}{{3q}}$
Putting these values in (3), we will get:-
$d = \dfrac{{1 - 6q}}{{3q}} - \dfrac{1}{{3q}}$
Taking LCM, we will get:-
$d = \dfrac{{1 - 6q - 1}}{{3q}}$
Simplifying it:
$d = \dfrac{{ - 6q}}{{3q}}$
Simplifying it further, we will have:-
$d = \dfrac{{ - 6}}{3} = - 2$
Hence, d = -2

Hence, the correct option is (D).

Note: The students must note that they may use any of the two consecutive terms to find the common difference. We here used first and second because they used less calculations than the second and third term. You many use any two. The students may use first and third but that would be incorrect as it will result in twice the common difference. So, if you try to use the first and third term, remember to divide the result by 2 to reach the final answer.