Verify that each of the following is an $AP$,and then write its next three terms.
$a, 2a+1, 3a+2, 4a+3, \ldots$
$a, 2a+1, 3a+2, 4a+3, \ldots$
(A) Given sequence: $a_1 = a, a_2 = 2a+1, a_3 = 3a+2, a_4 = 4a+3$.
Calculate the common difference $d$:
$d_1 = a_2 - a_1 = (2a+1) - a = a+1$
$d_2 = a_3 - a_2 = (3a+2) - (2a+1) = a+1$
$d_3 = a_4 - a_3 = (4a+3) - (3a+2) = a+1$
Since $d_1 = d_2 = d_3 = a+1$,the common difference is constant. Therefore,the sequence is an $AP$.
The next three terms are:
$a_5 = a_4 + d = (4a+3) + (a+1) = 5a+4$
$a_6 = a_5 + d = (5a+4) + (a+1) = 6a+5$
$a_7 = a_6 + d = (6a+5) + (a+1) = 7a+6$
Calculate the common difference $d$:
$d_1 = a_2 - a_1 = (2a+1) - a = a+1$
$d_2 = a_3 - a_2 = (3a+2) - (2a+1) = a+1$
$d_3 = a_4 - a_3 = (4a+3) - (3a+2) = a+1$
Since $d_1 = d_2 = d_3 = a+1$,the common difference is constant. Therefore,the sequence is an $AP$.
The next three terms are:
$a_5 = a_4 + d = (4a+3) + (a+1) = 5a+4$
$a_6 = a_5 + d = (5a+4) + (a+1) = 6a+5$
$a_7 = a_6 + d = (6a+5) + (a+1) = 7a+6$
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