The list of numbers $- 10, - 6, - 2, 2$ is ,

an AP with $d = - 16$

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an AP with $d = 4$

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an AP with $d = - 4$

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not an AP

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Solution

The correct option is B
an AP with $d = 4$

The explanation for the correct option.
A sequence will be an AP, when the common difference between two consecutive numbers will be the same.
If an AP is, $a_{1}, a_{2}, a_{3}, a_{4}, . . . . .$ . Then, the common difference $(d)$ is computed as,
$d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3}$
For the list of numbers, $- 10, - 6, - 2, 2$ , the common difference is computed as,
$d = - 6 - (- 10) = - 2 - (- 6) = 2 - (- 2) \Rightarrow d = - 6 + 10 = - 2 + 6 = 2 + 2 \Rightarrow d = 4 = 4 = 4$
Since the common difference between two consecutive numbers is the same.
Therefore, the given list of numbers are in AP with common differences of $4$ .
Hence, the option $B$ is correct.

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