Are $1, 3, 9, 27, \ldots$ in $AP$? If they form an $AP$,find the common difference $d$ and write three more terms.
(NONE) Given sequence: $1, 3, 9, 27, \ldots$
To check if the sequence is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = 3 - 1 = 2$
$a_{3} - a_{2} = 9 - 3 = 6$
$a_{4} - a_{3} = 27 - 9 = 18$
Since the common difference $(a_{k+1} - a_{k})$ is not constant (i.e.,$2 \neq 6 \neq 18$),the given sequence does not form an $AP$.
To check if the sequence is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = 3 - 1 = 2$
$a_{3} - a_{2} = 9 - 3 = 6$
$a_{4} - a_{3} = 27 - 9 = 18$
Since the common difference $(a_{k+1} - a_{k})$ is not constant (i.e.,$2 \neq 6 \neq 18$),the given sequence does not form an $AP$.
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| $(ii)$ | $-18$ | $...$ | $10$ | $0$ |
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