Question

Which of the following is true regarding the sequence below?

Answer 1

Answer:

Option D.

Step-by-step explanation:

The given sequence is $\frac{7}{12}, \frac{1}{6}, -\frac{1}{4}, \frac{2}{3}$

Now this sequence will be an arithmetic sequence if there is a common difference.

To check the common difference we will subtract each term from the successive term.

$T_{2}-T_{1}= \frac{1}{6}-\frac{7}{12}=-\frac{5}{12}$

$T_{3}-T_{2}=-\frac{1}{4}-\frac{1}{6}=-\frac{5}{12}$

$T_{4}-T_{3}=\frac{2}{3}+\frac{1}{4}=\frac{11}{12}$

Therefore $T_{2}-T_{1}=T_{3}-T_{2}\neq T_{4}-T_{3}$

So the answer is option D.The sequence is not an arithmetic sequence because there is no common difference between $a_{n}-a_{n-1}$

Answer 2

Answer: Option D

Explanation:

Arithmetic Progression is the sequence of numbers in which consecutive terms have same or common difference.

common difference d= a2-a1

Option A is incorrect because in last two terms there is not a common difference of $\frac{-5}{12}$

$\frac{1}{6} -\frac{7}{12} =\frac{-5}{12}$ , $\frac{-1}{4} -\frac{1}{6} =\frac{-5}{12}$ , $\frac{2}{3} -\frac{-1}{4} =\frac{11}{12}$.

Option B is incorrect because there is not a common difference of 1 in any of the given terms

Option C is incorrect because we do not check common difference in the number of terms that is n we check it for $a_{n}$.