$-10, -6, -2, 2, \ldots$ are $APs$? If they form an $AP$,find the common difference $d$ and write three more terms.
(D) Given sequence: $-10, -6, -2, 2, \ldots$
To check if it is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = (-6) - (-10) = -6 + 10 = 4$
$a_{3} - a_{2} = (-2) - (-6) = -2 + 6 = 4$
$a_{4} - a_{3} = 2 - (-2) = 2 + 2 = 4$
Since the difference $a_{k+1} - a_{k}$ is constant $(d = 4)$,the sequence forms an $AP$.
The next three terms are:
$a_{5} = a_{4} + d = 2 + 4 = 6$
$a_{6} = a_{5} + d = 6 + 4 = 10$
$a_{7} = a_{6} + d = 10 + 4 = 14$
Thus,the common difference is $d = 4$ and the next three terms are $6, 10, 14$.
To check if it is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = (-6) - (-10) = -6 + 10 = 4$
$a_{3} - a_{2} = (-2) - (-6) = -2 + 6 = 4$
$a_{4} - a_{3} = 2 - (-2) = 2 + 2 = 4$
Since the difference $a_{k+1} - a_{k}$ is constant $(d = 4)$,the sequence forms an $AP$.
The next three terms are:
$a_{5} = a_{4} + d = 2 + 4 = 6$
$a_{6} = a_{5} + d = 6 + 4 = 10$
$a_{7} = a_{6} + d = 10 + 4 = 14$
Thus,the common difference is $d = 4$ and the next three terms are $6, 10, 14$.
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